2.2 Solved examples
| | | | | |
| # |
original ode |
\(y=xf\left ( p\right ) +g\left ( p\right ) \) |
\(f\left ( p\right ) \) |
\(g\left ( p\right ) \) |
type |
| | | | | |
| \(1\) |
\(x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1\) |
\(y=xp+\frac {1}{p}\) |
\(p\) |
\(\frac {1}{p}\) |
Clairaut |
| | | | | |
| \(2\) |
\(y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}\) |
\(y=xp-p^{2}\) |
\(p\) |
\(-p^{2}\) |
Clairaut |
| | | | | |
| \(3\) |
\(y=xy^{\prime }-\frac {1}{4}\left ( y^{\prime }\right ) ^{2}\) |
\(y=xp-\frac {1}{4}p^{2}\) |
\(p\) |
\(-\frac {1}{4}p^{2}\) |
Clairaut |
| | | | | |
| \(4\) |
\(y=x\left ( y^{\prime }\right ) ^{2}\) |
\(y=xp^{2}\) |
\(p^{2}\) |
\(0\) |
d’Alembert |
| | | | | |
| \(5\) |
\(y=x+\left ( y^{\prime }\right ) ^{2}\) |
\(y=x+p^{2}\) |
\(1\) |
\(p^{2}\) |
d’Alembert |
| | | | | |
| \(6\) |
\(\left ( y^{\prime }\right ) ^{2}-1-x-y=0\) |
\(y=-x+\left ( p^{2}-1\right ) \) |
\(-1\) |
\(\left ( p^{2}-1\right ) \) |
d’Alembert |
| | | | | |
| \(7\) |
\(yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x\) |
\(y=\frac {1}{p}x+p\) |
\(\frac {1}{p}\) |
\(p\) |
d’Alembert |
| | | | | |
| \(8\) |
\(y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}\) |
\(y=xp^{2}+p^{2}\) |
\(p^{2}\) |
\(p^{2}\) |
d’Alembert |
| | | | | |
| \(9\) |
\(y=\frac {x}{a}y^{\prime }+\frac {b}{ay^{\prime }}\) |
\(y=\frac {x}{a}p+\frac {b}{a}\frac {1}{p}\) |
\(\frac {p}{a}\) |
\(\frac {b}{a}\frac {1}{p}\) |
d’Alembert |
| | | | | |
| \(10\) |
\(y=x\left ( y^{\prime }+a\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\right ) \) |
\(y=x\left ( p+a\sqrt {1+p^{2}}\right ) \) |
\(p+a\sqrt {1+p^{2}}\) |
\(0\) |
d’Alembert |
| | | | | |
| \(11\) |
\(y=x+\left ( y^{\prime }\right ) ^{2}\left ( 1-\frac {2}{3}y^{\prime }\right ) \) |
\(y=x+p^{2}\left ( 1-\frac {2}{3}p\right ) \) |
\(1\) |
\(p^{2}\left ( 1-\frac {2}{3}p\right ) \) |
d’Alembert |
| | | | | |
| \(12\) |
\(y=2x-\frac {1}{2}\ln \left ( \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\right ) \) | \(y=2x-\frac {1}{2}\ln \left ( \frac {p^{2}}{p-1}\right ) \) | \(2\) | \(-\frac {1}{2}\ln \left ( \frac {p^{2}}{p-1}\right ) \) | d’Alembert |
| | | | | |
| \(13\) | \(\left ( y^{\prime }\right ) ^{2}-x\left ( y^{\prime }\right ) ^{2}+y\left ( 1+y^{\prime }\right ) -xy^{\prime }=0\) | \(y=\frac {xp+xp^{2}-p^{2}}{p+1}=xp-\frac {p^{2}}{p+1}\) | \(p\) | \(-\frac {p^{2}}{p+1}\) | Clairaut |
| | | | | |
| \(14\) |
\(x\left ( y^{\prime }\right ) ^{2}+\left ( x-y\right ) y^{\prime }+1-y=0\) |
\(y=xp+\frac {1}{1+p}\) |
\(p\) |
\(\frac {1}{1+p}\) |
Clairaut |
| | | | | |
| \(15\) |
\(xyy^{\prime }=y^{2}+x\sqrt {4x^{2}+y^{2}}\) |
\(y=\operatorname {RootOf}\left ( h(p)\right ) x\) |
\(\operatorname {RootOf}\left ( h(p)\right ) \) |
\(0\) |
d’Alembert |
| | | | | |
| \(16\) |
\(\ln \left ( \cos y^{\prime }\right ) +y^{\prime }\tan y^{\prime }=y\) |
\(y=\ln \left ( \cos p\right ) +p\tan p\) |
\(0\) |
\(\ln \left ( \cos p\right ) +p\tan p\) |
d’Alembert |
| | | | | |
| \(17\) |
\(x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x=0\) |
\(y=x\left ( \frac {1}{2}p+\frac {2}{p}\right ) \) |
\(\frac {1}{2}p+\frac {2}{p}\) |
\(0\) |
d’Alembert |
| | | | | |
| \(18\) |
\(x-yy^{\prime }=a\left ( y^{\prime }\right ) ^{2}\) |
\(y=\frac {x}{p}-ap\) |
\(\frac {1}{p}\) |
\(-ap\) |
d’Alembert |
| | | | | |
| \(19\) |
\(y=xF\left ( p\right ) +G\left ( p\right ) \) |
\(y=xF\left ( p\right ) +G\left ( p\right ) \) |
\(F\left ( p\right ) \) |
\(G\left ( p\right ) \) |
d’Alembert |
| | | | | |
| \(20\) |
\(y^{\prime }=-\frac {x}{2}-1+\frac {1}{2}\sqrt {x^{2}+4x+4y}\) |
\(y=xp+\left ( 1+2p+p^{2}\right ) \) |
\(p\) |
\(1+2p+p^{2}\) |
Clairaut |
| | | | | |
| \(21\) |
\(\frac {y^{\prime }y}{1+\frac {1}{2}\sqrt {1+\left ( y^{\prime }\right ) ^{2}}}=-x\) |
\(y=-x\left ( \frac {2+\sqrt {1+p^{2}}}{2p}\right ) \) |
\(-\left ( \frac {2+\sqrt {1+p^{2}}}{2p}\right ) \) |
\(0\) |
d’Alembert |
| | | | | |
| \(22\) |
\(x\left ( y^{\prime }\right ) ^{3}=yy^{\prime }+1\) |
\(y=xp^{2}-\frac {1}{p}\) |
\(p^{2}\) |
\(-\frac {1}{p}\) |
d’Alembert |
| | | | | |
| \(23\) |
\(\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }=2x\) |
\(y=-x\frac {1}{p}+\frac {1}{2}p\) |
\(-\frac {1}{p}\) |
\(\frac {1}{2}p\) |
d’Alembert |
| | | | | |
| \(24\) |
\(xy^{\prime }-y=\sqrt {x^{2}-y^{2}}\) |
\(y=x\left ( \frac {p}{2}\pm \frac {1}{2}\sqrt {2-p^{2}}\right ) \) |
\(\frac {p}{2}\pm \frac {1}{2}\sqrt {2-p^{2}}\) |
\(0\) |
d’Alembert |
| | | | | |
2.2.1 Example 1
\(x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }=-1\), is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives
\begin{align} y & =xp+\frac {1}{p}\tag {1}\\ & =xf\left ( p\right ) +g\left ( p\right ) \nonumber \end{align}
Where \(f\left ( p\right ) =p\) and \(g\left ( p\right ) =\frac {1}{p}\). Since \(f\left ( p\right ) =p\) then this is Clairaut ode. Taking derivative of the above w.r.t. \(x\)
gives
\begin{align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\end{align*}
The general solution is given by
\begin{align*} \frac {dp}{dx} & =0\\ p & =c_{1}\end{align*}
Substituting this in (1) gives the general solution
\[ y=c_{1}x+\frac {1}{c_{1}}\]
The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular
solutions.
\begin{align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\frac {1}{p}\\ & =x-\frac {1}{p^{2}}\end{align*}
Hence \(x-\frac {1}{p^{2}}=0\) or \(p=\pm \frac {1}{\sqrt {x}}\). Substituting these back in (1) gives
\begin{align} y_{1}\left ( x\right ) & =xp+\frac {1}{p}\nonumber \\ & =x\frac {1}{\sqrt {x}}+\sqrt {x}\nonumber \\ & =2\sqrt {x}\tag {3}\\ y_{2}\left ( x\right ) & =-x\sqrt {\frac {1}{x}}-\sqrt {x}\nonumber \\ & =-2\sqrt {x} \tag {4}\end{align}
Eq. (2) is the general solution and (3,4) are the singular solutions.
Another method to find the singular solutions if it exists is called the p-discriminant. This
is used only for first order ode with nonlinear in \(y^{\prime }\). We set up the following two
equations
\begin{align*} F\left ( x,y,y^{\prime }\right ) & =0\\ \frac {\partial F\left ( x,y,y^{\prime }\right ) }{\partial y^{\prime }} & =0 \end{align*}
We eliminate \(y^{\prime }\) and obtain \(G\left ( x,y\right ) =0\) equation. This is the singular solution. But we still have to check
if it satisfies the ode and also if it is true singular solution curve. More on this later. Let us
now just find the singular solution found above but using the p-discriminant method. The
above two equations are
\begin{align*} y-xy^{\prime }-\frac {1}{y^{\prime }} & =0\\ -x+\frac {1}{\left ( y^{\prime }\right ) ^{2}} & =0 \end{align*}
Second equation gives \(\left ( y^{\prime }\right ) ^{2}=\frac {1}{x}\). Hence \(y^{\prime }=\pm \sqrt {\frac {1}{x}}\). Hence the first equation now gives (starting with positive
root)
\begin{align*} y-x\sqrt {\frac {1}{x}}-\frac {1}{\sqrt {\frac {1}{x}}} & =0\\ y & =x\sqrt {\frac {1}{x}}+\frac {1}{\sqrt {\frac {1}{x}}}\\ & =\frac {x\sqrt {\frac {1}{x}}\sqrt {\frac {1}{x}}+1}{\sqrt {\frac {1}{x}}}\\ & =2\sqrt {x}\end{align*}
And for the second root \(y^{\prime }=-\sqrt {\frac {1}{x}}\) we obtain \(y=-2\sqrt {x}\). We see these are the same singular solutions obtained
earlier.
2.2.2 Example 2
\(y=xy^{\prime }-\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives
\begin{align} y & =xp-p^{2}\tag {1}\\ & =xf\left ( p\right ) +g\left ( p\right ) \nonumber \end{align}
Where \(f\left ( p\right ) =p\) and \(g\left ( p\right ) =-p^{2}\). Since \(f\left ( p\right ) =p\) then this is Clairaut ode. Taking derivative of the above w.r.t. \(x\)
gives
\begin{align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\end{align*}
The general solution is given by
\begin{align*} \frac {dp}{dx} & =0\\ p & =c_{1}\end{align*}
Substituting this in (1) gives the general solution
\[ y=c_{1}x-c_{1}^{2}\]
The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular
solutions.
\begin{align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\left ( -p^{2}\right ) \\ & =x+2p \end{align*}
Hence \(x+2p=0\) or \(p=\frac {x}{2}\). Substituting this back in (1) gives
\begin{align} y\left ( x\right ) & =\frac {x^{2}}{2}-\frac {x^{2}}{4}\nonumber \\ & =\frac {x^{2}}{4} \tag {3}\end{align}
Eq. (2) is the general solution and (3) is the singular solution.
Another method to find the singular solutions if it exists is called the p-discriminant. This
is used only for first order ode with nonlinear in \(y^{\prime }\). We set up the following two
equations
\begin{align*} F\left ( x,y,y^{\prime }\right ) & =0\\ \frac {\partial F\left ( x,y,y^{\prime }\right ) }{\partial y^{\prime }} & =0 \end{align*}
We eliminate \(y^{\prime }\) and obtain \(G\left ( x,y\right ) =0\) equation. This is the singular solution. But we still have to check
if it satisfies the ode and also if it is true singular solution curve. More on this later. Let us
now just find the singular solution found above but using the p-discriminant method. The
above two equations are
\begin{align*} y-xy^{\prime }+\left ( y^{\prime }\right ) ^{2} & =0\\ -x+2y^{\prime } & =0 \end{align*}
Second equation gives \(y^{\prime }=\frac {x}{2}\). Hence the first equation now gives the singular solution as
\begin{align*} y-x\left ( \frac {x}{2}\right ) +\left ( \frac {x}{2}\right ) ^{2} & =0\\ y & =\frac {x^{2}}{2}-\frac {x^{2}}{4}\\ & =\frac {1}{4}x^{2}\end{align*}
Which is the same obtained earlier.
2.2.3 Example 3
\(y=xy^{\prime }-\frac {1}{4}\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives
\begin{align} y & =xp-\frac {1}{4}p^{2}\tag {1}\\ & =xf\left ( p\right ) +g\left ( p\right ) \nonumber \end{align}
Where \(f\left ( p\right ) =p\) and \(g\left ( p\right ) =-\frac {1}{4}p^{2}\). Since \(f\left ( p\right ) =p\) then this is Clairaut ode. Taking derivative of the above w.r.t. \(x\)
gives
\begin{align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\end{align*}
The general solution is given by
\begin{align*} \frac {dp}{dx} & =0\\ p & =c_{1}\end{align*}
Substituting this in (1) gives the general solution
\[ y=c_{1}x-\frac {1}{4}c_{1}^{2}\]
The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular
solutions.
\begin{align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\left ( -\frac {1}{4}p^{2}\right ) \\ & =x-\frac {1}{2}p \end{align*}
Hence \(x-\frac {1}{2}p=0\) or \(p=2x\). Substituting this back in (1) gives
\begin{align} y\left ( x\right ) & =2x^{2}-x^{2}\nonumber \\ & =x^{2} \tag {3}\end{align}
Eq. (2) is the general solution and (3) is the singular solution.
Another method to find the singular solutions if it exists is called the p-discriminant. This
is used only for first order ode with nonlinear in \(y^{\prime }\). We set up the following two
equations
\begin{align*} F\left ( x,y,y^{\prime }\right ) & =0\\ \frac {\partial F\left ( x,y,y^{\prime }\right ) }{\partial y^{\prime }} & =0 \end{align*}
We eliminate \(y^{\prime }\) and obtain \(G\left ( x,y\right ) =0\) equation. This is the singular solution. But we still have to check
if it satisfies the ode and also if it is true singular solution curve. More on this later. Let us
now just find the singular solution found above but using the p-discriminant method. The
above two equations are
\begin{align*} y-xy^{\prime }+\frac {1}{4}\left ( y^{\prime }\right ) ^{2} & =0\\ -x+\frac {1}{2}y^{\prime } & =0 \end{align*}
Second equation gives \(y^{\prime }=2x\). Hence the first equation now gives the singular solution as
\begin{align*} y-2x^{2}+\frac {1}{4}\left ( 4x^{2}\right ) & =0\\ y-x^{2} & =0\\ y & =x^{2}\end{align*}
Which is the same obtained earlier.
2.2.4 Example 4
\(y=x\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) and solving for \(y\) gives
\begin{align} y & =xp^{2}\tag {1}\\ & =xf\left ( p\right ) \nonumber \end{align}
This is the case when \(f\left ( p\right ) =p^{2}\) and \(g\left ( p\right ) =0\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode.
Writing \(f\equiv f\left ( p\right ) \) and \(g\equiv g\left ( p\right ) \) to make notation simpler but remembering that \(f\) is function of \(p\left ( x\right ) \)
which in turn is function of \(x\). Same for \(g\left ( p\right ) \).
\[ y=xf \]
Taking derivative of the above w.r.t. \(x\)
gives
\begin{align*} p & =\frac {d}{dx}\left ( xf\right ) \\ p & =f+xf^{\prime }\frac {dp}{dx}\\ p-f & =xf^{\prime }\frac {dp}{dx}\end{align*}
Since \(f=p^{2}\) then the above becomes
\begin{equation} p-p^{2}=2xp\frac {dp}{dx} \tag {2}\end{equation}
The singular solution is given when \(\frac {dp}{dx}=0\) or \(p-p^{2}=0\). This gives \(p=0\) or \(p=1\).
Substituting these values of \(p\) in (1) gives singular solutions
\begin{align} y_{s1} & =0\tag {3}\\ y_{s2} & =x \tag {4}\end{align}
General solution is found when \(\frac {dp}{dx}\neq 0\,\). Eq(2) is a first order ode in \(p\). Now we could either solve ode
(2) directly as it is for \(p\left ( x\right ) \), or do an inversion and solve for \(x\left ( p\right ) \). If the ode is linear as is in \(p\) then
no need to do inversion. Since (2) is separable as is, no need to do an inversion. The solution
to (2) is
\begin{align*} p_{1} & =0\\ p_{2} & =1+\frac {c_{1}}{\sqrt {x}}\end{align*}
For each \(p\), there is a general solution. Substituting each of the above in (1) gives
\begin{align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2}\end{align*}
Hence the final solutions are
\begin{align*} y & =x\qquad \text {(singular)}\\ y & =0\\ y & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2}\end{align*}
But \(y=x\) can be obtained from the general solution when \(c_{1}=0\). Hence it is removed. Therefore the
final solutions are
\begin{align} y & =0\tag {6}\\ y & =x\left ( 1+\frac {c_{1}}{\sqrt {x}}\right ) ^{2} \tag {7}\end{align}
What will happen if we had done an inversion to \(x\left ( p\right ) \)? Let us find out. ode(5) now
becomes
\begin{align*} \frac {p-p^{2}}{p}\frac {dx}{dp} & =2x\\ \frac {dx}{2x} & =\frac {p}{p-p^{2}}dp \end{align*}
This is also separable in \(x\). Solving this for \(x\) gives
\[ x=\frac {c_{1}}{\left ( p-1\right ) ^{2}}\]
Solving for \(p\) from the above gives
\begin{align*} p_{1} & =\frac {x+\sqrt {xc_{1}}}{x}\\ p_{2} & =\frac {x-\sqrt {xc_{1}}}{x}\end{align*}
Substituting each of the above in (1) gives
\begin{align*} y_{1} & =x\left ( \frac {x+\sqrt {xc_{1}}}{x}\right ) ^{2}\\ & =\frac {\left ( x+\sqrt {xc_{1}}\right ) }{x}^{2}\\ y_{2} & =x\left ( \frac {x-\sqrt {xc_{1}}}{x}\right ) ^{2}\\ & =\frac {\left ( x-\sqrt {xc_{1}}\right ) }{x}^{2}\end{align*}
Now we see that singular solution \(y=x\) can be obtained from the above general solutions from \(c_{1}=0\).
But \(y=0\) can not. Hence the final solutions are
\begin{align} y & =0\qquad \text {(singular)}\tag {8}\\ y & =\frac {\left ( x+\sqrt {xc_{1}}\right ) }{x}^{2}\tag {9}\\ y & =\frac {\left ( x-\sqrt {xc_{1}}\right ) }{x}^{2} \tag {10}\end{align}
All solutions (6,7,8,9,10) are correct and verified. Maple gives the solutions given in (8,9,10)
and not those in (6,7).
Another method to find the singular solutions if it exists is called the p-discriminant. This
is used only for first order ode with nonlinear in \(y^{\prime }\). We set up the following two
equations
\begin{align*} F\left ( x,y,y^{\prime }\right ) & =0\\ \frac {\partial F\left ( x,y,y^{\prime }\right ) }{\partial y^{\prime }} & =0 \end{align*}
We eliminate \(y^{\prime }\) and obtain \(G\left ( x,y\right ) =0\) equation. This is the singular solution. But we still have to check
if it satisfies the ode and also if it is true singular solution curve. More on this later. Let us
now just find the singular solution found above but using the p-discriminant method. The
above two equations are
\begin{align*} y-x\left ( y^{\prime }\right ) ^{2} & =0\\ -2xy^{\prime } & =0 \end{align*}
Second equation gives \(y^{\prime }=0\). Hence the first equation now gives the singular solution as
\[ y=0 \]
Which is
the same obtained earlier.
2.2.5 Example 5
\(y=x+\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives
\begin{align} y & =x+p^{2}\tag {1}\\ & =xf+g\nonumber \end{align}
Hence \(f\left ( p\right ) =1,g\left ( p\right ) =p^{2}\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives
\begin{align} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \nonumber \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \tag {2}\end{align}
Using \(f=1,g=p^{2}\) the above simplifies to
\begin{equation} p-1=2p\frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in (2) which results in \(p-f=0\) or \(p-1=0\). Hence \(p=1\).
Substituting these values of \(p\) in (1) gives singular solution as
\begin{equation} y=x+1 \tag {3}\end{equation}
General solution is
found when \(\frac {dp}{dx}\neq 0\,\). Eq (2A) is a first order ode in \(p\). Now we could either solve ode (2)
directly as it is for \(p\left ( x\right ) \), or do an inversion and solve for \(x\left ( p\right ) \). Since (2) is separable as is,
no need to do an inversion. Solving (2) for \(p\) gives
\[ p=\operatorname *{LambertW}\left ( c_{1}e^{\frac {x}{2}-1}\right ) +1 \]
Substituting this in (1) gives
the general solution
\begin{equation} y\left ( x\right ) =x+\left ( \operatorname *{LambertW}\left ( c_{1}e^{\frac {x}{2}-1}\right ) +1\right ) ^{2} \tag {4}\end{equation}
Note however that when \(c_{1}=0\) then the general solution becomes
\(y\left ( x\right ) =x+1\). Hence (3) is a particular solution and not a singular solution. (4) is the only
solution.
2.2.6 Example 6
\(\left ( y^{\prime }\right ) ^{2}-1-x-y=0\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives
\begin{align} y & =-x+\left ( p^{2}-1\right ) \tag {1}\\ & =xf+g\nonumber \end{align}
Hence \(f=-1,g\left ( p\right ) =\left ( p^{2}-1\right ) \). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives
\begin{align} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \nonumber \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \tag {2}\end{align}
Using \(f=-1,g=\left ( p^{2}-1\right ) \) the above simplifies to
\begin{equation} p+1=2p\frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in
\(p=-1\). Substituting this in (1) gives singular solution as
\begin{equation} y\left ( x\right ) =-x \tag {3}\end{equation}
The general solution is found by finding \(p\)
from (2A). No need here to do the inversion as (2) is separable already. Solving (2)
gives
\begin{align*} p & =-\operatorname *{LambertW}\left ( -e^{-\frac {x}{2}-1+\frac {c_{2}}{2}}\right ) -1\\ & =-\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1 \end{align*}
Substituting the above in (1) gives the general solution
\begin{align} y\left ( x\right ) & =-x+\left ( p^{2}-1\right ) \nonumber \\ y\left ( x\right ) & =-x+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{-\frac {x}{2}-1}\right ) -1\right ) ^{2}-1 \tag {4}\end{align}
Note however that when \(c_{1}=0\) then the general solution becomes \(y\left ( x\right ) =-x\). Hence (3) is a particular
solution and not a singular solution. Solution (4) is therefore the only solution.
2.2.7 Example 7
\(yy^{\prime }-\left ( y^{\prime }\right ) ^{2}=x\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives
\begin{align} y & =\frac {x+p^{2}}{p}\tag {1}\\ & =\frac {1}{p}x+p\nonumber \\ & =xf+g\nonumber \end{align}
Hence \(f=\frac {1}{p},g\left ( p\right ) =p\). Taking derivative w.r.t. \(x\) gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}
Using \(f=\frac {1}{p},g=p\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. the above simplifies to
\begin{equation} p-\frac {1}{p}=\left ( -\frac {x}{p^{2}}+1\right ) \frac {dp}{dx} \tag {2A}\end{equation}
The singular
solution is found by setting \(\frac {dp}{dx}=0\) in (2) which results in \(Q\left ( p\right ) =0\) or \(p-1=0\) or \(p=1\). Substituting these values
in (1) gives the solutions
\begin{equation} y_{1}\left ( x\right ) =x+1 \tag {3}\end{equation}
The general solution is found by finding \(p\) from (2A).
Since (2A) is not linear and not separable in \(p\), then inversion is needed. Writing (2)
as
\begin{align*} \frac {dx}{dp} & =\frac {1-\frac {x}{p^{2}}}{p-\frac {1}{p}}\\ & =\frac {1}{p-p^{3}}\left ( x-p^{2}\right ) \end{align*}
Hence
\[ \frac {dx}{dp}+\frac {x}{p\left ( p^{2}-1\right ) }=\frac {p^{2}}{p\left ( p^{2}-1\right ) }\]
This is now linear ODE in \(x\left ( p\right ) \). The solution is
\begin{align} x & =\frac {p\sqrt {\left ( p-1\right ) \left ( 1+p\right ) }\ln \left ( p+\sqrt {p^{2}-1}\right ) }{\left ( 1+p\right ) \left ( p-1\right ) }+c_{1}\frac {p}{\sqrt {\left ( 1+p\right ) \left ( p-1\right ) }}\nonumber \\ & =\frac {p\sqrt {p^{2}-1}\ln \left ( p+\sqrt {p^{2}-1}\right ) }{p^{2}-1}+c_{1}\frac {p}{\sqrt {p^{2}-1}} \tag {4}\end{align}
Now we need to eliminate \(p\) from (1,4). From (1) since \(y=\frac {1}{p}x+p\) then solving for \(p\) gives
\begin{align*} p_{1} & =\frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\\ p_{2} & =\frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\end{align*}
Substituting each \(p_{i}\) in (4) gives the general solution (implicit) in \(y\left ( x\right ) \). First solution is
\[ x=\frac {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) \sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}+\sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac {\frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}}{\sqrt {\left ( \frac {y}{2}+\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}}\]
And second
solution is
\[ x=\frac {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) \sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\ln \left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}+\sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}\right ) }{\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}+c_{1}\frac {\frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}}{\sqrt {\left ( \frac {y}{2}-\frac {1}{2}\sqrt {y^{2}-4x}\right ) ^{2}-1}}\]
2.2.8 Example 8
\(y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives
\begin{align} y & =xp^{2}+p^{2}\tag {1}\\ & =xf+g\nonumber \end{align}
where \(f=p^{2},g=p^{2}\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative and simplifying gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}
Using values for \(f,g\) the above simplifies to
\begin{equation} p-p^{2}=\left ( 2xp+2p\right ) \frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which
results in \(p=0\) or \(p=1\). Substituting these values in (1) gives the singular solutions
\begin{align} y_{1}\left ( x\right ) & =0\tag {3}\\ y_{2}\left ( x\right ) & =x+1 \tag {4}\end{align}
The general solution is found by finding \(p\) from (2A). Since (2A) is not linear in \(p\), then
inversion is needed. Writing (A2) as
\[ \frac {p\left ( 1-p\right ) }{2p\left ( x+1\right ) }=\frac {dp}{dx}\]
Inverting gives
\begin{align*} \frac {dx}{dp} & =\frac {2\left ( x+1\right ) }{\left ( 1-p\right ) }\\ \frac {dx}{dp}-x\frac {2}{\left ( 1-p\right ) } & =\frac {2}{\left ( 1-p\right ) }\end{align*}
This is now linear \(x\left ( p\right ) \). The solution is
\[ x=\frac {C^{2}}{\left ( p-1\right ) ^{2}}-1 \]
Solving for \(p\) gives
\begin{align*} \frac {C^{2}}{\left ( p-1\right ) ^{2}} & =x+1\\ \left ( p-1\right ) ^{2} & =\frac {C^{2}}{x+1}\\ \left ( p-1\right ) & =\pm \frac {C}{\sqrt {x+1}}\\ p & =1\pm \frac {C}{\sqrt {x+1}}\end{align*}
Substituting the above in (1) gives the general solutions
\[ y=\left ( x+1\right ) p^{2}\]
Therefore
\begin{align*} y\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac {C}{\sqrt {x+1}}\right ) ^{2}\\ y\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac {C}{\sqrt {x+1}}\right ) ^{2}\end{align*}
The solution \(y_{1}\left ( x\right ) =0\) found earlier can not be obtained from the above general solution hence it is
singular solution. But \(y_{2}\left ( x\right ) =x+1\) can be obtained from the general solution when \(C=0\). Hence there are
only three solutions, they are
\begin{align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac {C}{\sqrt {x+1}}\right ) ^{2}\\ y_{3}\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac {C}{\sqrt {x+1}}\right ) ^{2}\end{align*}
2.2.9 Example 9
\(y=\frac {x}{a}y^{\prime }+\frac {b}{ay^{\prime }}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives
\begin{align} y & =\frac {x}{a}p+\frac {b}{a}p^{-1}\tag {1}\\ & =xf+g\nonumber \end{align}
Where \(f=\frac {p}{a},g=\frac {b}{a}p^{-1}\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}
Using values for \(f,g\) the above simplifies to
\begin{equation} p-\frac {p}{a}=\left ( \frac {x}{a}-\frac {b}{a}p^{-2}\right ) \frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which
results in \(p=0\). Substituting this in (1) does not generate any solutions due to division by zero.
Hence no singular solution exist.
The general solution is found by finding \(p\) from (2A). Since (2A) is not linear in \(p\),
then inversion is needed. Writing (2A) as
\[ \frac {p\left ( 1-\frac {1}{a}\right ) }{\frac {x}{a}-\frac {b}{a}p^{-2}}=\frac {dp}{dx}\]
Since this is nonlinear, then inversion is
needed
\begin{align*} \frac {dx}{dp} & =\frac {\frac {x}{a}-\frac {b}{a}p^{-2}}{p\left ( 1-\frac {1}{a}\right ) }\\ \frac {dx}{dp}-x\frac {1}{p\left ( a-1\right ) } & =-\frac {b}{a}\frac {1}{p^{3}\left ( 1-\frac {1}{a}\right ) }\end{align*}
This is now linear ode in \(x\left ( p\right ) \). The solution is
\begin{equation} x=\frac {b}{(2a-1)p^{2}}+C_{1}p^{\frac {1}{a-1}} \tag {3}\end{equation}
There are now two choices to take. The first is by
solving for \(p\) from the above in terms of \(x\) and then substituting the result in (1) to obtain
explicit solution for \(y\left ( x\right ) \), and the second choice is by solving for \(p\) algebraically from (1) and
substituting the result in (3). The second choice is easier in this case but gives an implicit
solution. Solving for \(p\) from (1) gives
\begin{align*} p_{1} & =\frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\\ p_{1} & =\frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\end{align*}
Substituting each one of these solutions back in (3) gives two implicit solutions
\begin{align*} x & =\frac {b}{(2a-1)\left ( \frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{2}}+C_{1}\left ( \frac {ay+\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{\frac {1}{a-1}}\\ x & =\frac {b}{(2a-1)\left ( \frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{2}}+C_{1}\left ( \frac {ay-\sqrt {a^{2}y^{2}-4xb}}{2x}\right ) ^{\frac {1}{a-1}}\end{align*}
2.2.10 Example 10
\(y=xy^{\prime }+ax\sqrt {1+\left ( y^{\prime }\right ) ^{2}}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives
\begin{align} y & =x\left ( p+a\sqrt {1+p^{2}}\right ) \tag {1}\\ & =xf\nonumber \end{align}
where \(f=p+a\sqrt {1+p^{2}},g=0\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative and simplifying gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) \\ p-f & =xf^{\prime }\frac {dp}{dx}\end{align*}
Using values for \(f,g\) the above simplifies to
\begin{equation} -a\sqrt {1+p^{2}}=x\left ( 1+\frac {ap}{\sqrt {1+p^{2}}}\right ) \frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which
results in \(-a\sqrt {1+p^{2}}=0\). This gives no real solution for \(p.\) Hence no singular solution exists.
The general solution is when \(\frac {dp}{dx}\neq 0\) in (2A). Since (2A) is nonlinear, inversion is needed.
\begin{align*} \frac {-a\sqrt {1+p^{2}}}{x+\frac {1}{2}x\frac {2ap}{\sqrt {1+p^{2}}}} & =\frac {dp}{dx}\\ \frac {dx}{dp} & =\frac {x\left ( 1+\frac {1}{2}\frac {2ap}{\sqrt {1+p^{2}}}\right ) }{-a\sqrt {1+p^{2}}}\\ \frac {dx}{x} & =\frac {1+\frac {1}{2}\frac {2ap}{\sqrt {1+p^{2}}}}{-a\sqrt {1+p^{2}}}dp\\ \frac {dx}{x} & =\frac {\sqrt {1+p^{2}}+\frac {1}{2}2ap}{-a\left ( 1+p^{2}\right ) }dp\\ \frac {dx}{x} & =\left ( -\frac {1}{a\sqrt {1+p^{2}}}-\frac {p}{\left ( 1+p^{2}\right ) }\right ) dp \end{align*}
Integrating gives
\[ \ln x\left ( p\right ) =-\frac {1}{2}\ln \left ( p^{2}+1\right ) -\frac {1}{a}\operatorname {arcsinh}\left ( p\right ) \]
Therefore
\begin{equation} x=c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( p\right ) \right ) }}{\sqrt {p^{2}+1}} \tag {3}\end{equation}
There are now two choices to take. The first is by solving for \(p\)
from the above in terms of \(x\) and substituting the result in (1) to obtain explicit solution for \(y\left ( x\right ) \),
and the second choice is by solving for \(p\) algebraically from (1) and substituting the result in
(3). The second choice is easier in this case but gives an implicit solution. Solving for \(p\) from
(1) gives
\begin{align*} p_{1} & =-\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\\ p_{2} & =\frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\end{align*}
Substituting each one of these solutions back in (3) gives two implicit solutions
\begin{align*} x & =c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( -\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) \right ) }}{\sqrt {\left ( -\frac {1}{x}\frac {ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) ^{2}+1}}\\ x & =c_{1}\frac {-e^{-\frac {1}{a}\left ( \operatorname {arcsinh}\left ( \frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) \right ) }}{\sqrt {\left ( \frac {1}{x}\frac {-ay+\sqrt {-a^{2}x^{2}+x^{2}+y^{2}}a-y}{a^{2}-1}\right ) ^{2}+1}}\end{align*}
2.2.11 Example 11
\begin{align*} y & =x+\left ( y^{\prime }\right ) ^{2}\left ( 1-\frac {2}{3}y^{\prime }\right ) \\ & =x+p^{2}\left ( 1-\frac {2}{3}p\right ) \end{align*}
Where \(f=1,g=p^{2}\left ( 1-\frac {2}{3}p\right ) \). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}
Using values for \(f,g\) the above simplifies to
\begin{equation} p-1=\left ( 2p-2p^{2}\right ) \frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is when \(\frac {dp}{dx}=0\) which results in \(p=1\).
Substituting this in (1) gives
\begin{align*} y & =x-\left ( 1-\frac {2}{3}\right ) \\ & =x+\frac {1}{3}\end{align*}
The general solution is when \(\frac {dp}{dx}\neq 0\). Then (2A) is now separable. Solving for \(p\) gives
\begin{align*} p & =-\sqrt {c_{1}-x}\\ p & =\sqrt {c_{1}-x}\end{align*}
Substituting each one of the above solutions of \(p\) in (1) gives
\begin{align*} y_{1} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( -\sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( -\sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\end{align*}
And
\begin{align*} y_{2} & =x+\left ( p^{2}-\frac {2}{3}p^{3}\right ) \\ & =x+\left ( \left ( \sqrt {c_{1}-x}\right ) ^{2}-\frac {2}{3}\left ( \sqrt {c_{1}-x}\right ) ^{3}\right ) \\ & =x+\left ( c_{1}-x-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\right ) \\ & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\end{align*}
Therefore the solutions are
\begin{align*} y & =x+\frac {1}{3}\\ y & =c_{1}+\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\\ y & =c_{1}-\frac {2}{3}\left ( c_{1}-x\right ) ^{\frac {3}{2}}\end{align*}
2.2.12 Example 12
\begin{align*} \left ( y^{\prime }\right ) ^{2} & =e^{4x-2y}\left ( y^{\prime }-1\right ) \\ \ln \left ( y^{\prime }\right ) ^{2} & =\left ( 4x-2y\right ) +\ln \left ( y^{\prime }-1\right ) \\ 4x-2y & =\ln \left ( y^{\prime }\right ) ^{2}-\ln \left ( y^{\prime }-1\right ) \\ 4x-2y & =\ln \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\\ 2y & =4x-\ln \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\\ y & =2x-\frac {1}{2}\ln \left ( \frac {\left ( y^{\prime }\right ) ^{2}}{y^{\prime }-1}\right ) \\ & =2x-\frac {1}{2}\ln \left ( \frac {p^{2}}{p-1}\right ) \\ & =xf+g \end{align*}
Where \(f=2,g=-\frac {1}{2}\ln \left ( \frac {p^{2}}{p-1}\right ) \). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative w.r.t. \(x\) gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}
Using values for \(f,g\) the above simplifies to
\begin{equation} p-2=\left ( \frac {2-p}{2p^{2}-2p}\right ) \frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is when \(\frac {dp}{dx}=0\) which gives \(p=2\). From
(1) this gives
\[ y=2x-\frac {1}{2}\ln 4 \]
The general solution is when \(\frac {dp}{dx}\neq 0\). Then (2) becomes
\begin{align*} \frac {dp}{dx} & =\left ( p-2\right ) \left ( \frac {2p^{2}-2p}{2-p}\right ) \\ & =2p\left ( 1-p\right ) \end{align*}
is now separable. Solving for \(p\) gives
\[ p=\frac {1}{1+ce^{-2x}}\]
Substituting the above solutions of \(p\) in (1) gives
\begin{align*} y & =2x-\frac {1}{2}\ln \left ( \frac {\left ( \frac {1}{1+ce^{-2x}}\right ) ^{2}}{\frac {1}{1+ce^{-2x}}-1}\right ) \\ & =2x-\frac {1}{2}\ln \left ( \frac {-e^{4x}}{c\left ( c+e^{2x}\right ) }\right ) \end{align*}
2.2.13 Example 13
\begin{align} y & =\frac {xy^{\prime }+x\left ( y^{\prime }\right ) ^{2}-\left ( y^{\prime }\right ) ^{2}}{y^{\prime }+1}\nonumber \\ & =\frac {xp+xp^{2}-p^{2}}{p+1}\nonumber \\ & =xp-\frac {p^{2}}{p+1}\tag {1}\\ & =xf+g\nonumber \end{align}
Where \(f=p\) and \(g=-\frac {p^{2}}{p+1}\). Since \(f\left ( p\right ) =p\) then this is Clairaut ode. Taking derivative of the above w.r.t. \(x\)
gives
\begin{align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\end{align*}
The general solution is given by
\begin{align*} \frac {dp}{dx} & =0\\ p & =c_{1}\end{align*}
Substituting this in (1) gives the general solution
\[ y=xc_{1}-\frac {c_{1}^{2}}{c_{1}+1}\]
The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular
solutions.
\begin{align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\frac {1}{p}\\ & =x-\frac {1}{p^{2}}\end{align*}
Hence \(x-\frac {1}{p^{2}}=0\) or \(p=\pm \frac {1}{\sqrt {x}}\). Substituting these back in (1) gives
\begin{align} y_{1}\left ( x\right ) & =xp+\frac {1}{p}\nonumber \\ & =x\frac {1}{\sqrt {x}}+\sqrt {x}\nonumber \\ & =2\sqrt {x}\tag {3}\\ y_{2}\left ( x\right ) & =-x\sqrt {\frac {1}{x}}-\sqrt {x}\nonumber \\ & =-2\sqrt {x} \tag {4}\end{align}
Eq. (2) is the general solution and (3,4) are the singular solutions.
2.2.14 Example 14
\begin{align*} x\left ( y^{\prime }\right ) ^{2}+\left ( x-y\right ) y^{\prime }+1-y & =0\\ x\left ( y^{\prime }\right ) ^{2}+xy^{\prime }-yy^{\prime }+1-y & =0\\ y\left ( -y^{\prime }-1\right ) +x\left ( y^{\prime }\right ) ^{2}+xy^{\prime }+1 & =0 \end{align*}
Solving for \(y\)
\begin{align} y & =\frac {-x\left ( y^{\prime }\right ) ^{2}-xy^{\prime }-1}{-y^{\prime }-1}\nonumber \\ & =\frac {-xp^{2}-xp-1}{-p-1}\nonumber \\ & =\frac {xp^{2}+xp+1}{p+1}\nonumber \\ & =x\left ( \frac {p^{2}+p}{p+1}\right ) +\frac {1}{1+p}\nonumber \\ & =xp+\frac {1}{1+p}\nonumber \\ & =xf+g \tag {1}\end{align}
Where \(f=p\) and \(g=\frac {1}{1+p}\). Since \(f\left ( p\right ) =p\) then this is Clairaut ode. Taking derivative of the above w.r.t. \(x\)
gives
\begin{align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\end{align*}
The general solution is given by
\begin{align*} \frac {dp}{dx} & =0\\ p & =c_{1}\end{align*}
Substituting this in (1) gives the general solution
\begin{equation} y=c_{1}x+\frac {1}{c_{1}+1} \tag {4}\end{equation}
The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular
solutions. But
\begin{align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\left ( \frac {1}{1+p}\right ) \\ & =x-\frac {1}{\left ( p+1\right ) ^{2}}\end{align*}
Hence
\begin{align*} x-\frac {1}{\left ( p+1\right ) ^{2}} & =0\\ x\left ( p+1\right ) ^{2}-1 & =0\\ \left ( p+1\right ) ^{2} & =\frac {1}{x}\\ p+1 & =\pm \frac {1}{\sqrt {x}}\\ p & =\pm \frac {1}{\sqrt {x}}-1 \end{align*}
Substituting these values into (1) gives
\begin{align} y_{1} & =xp_{1}+\frac {1}{1+p_{1}}\nonumber \\ & =x\left ( \frac {1}{\sqrt {x}}-1\right ) +\frac {1}{1+\left ( \frac {1}{\sqrt {x}}-1\right ) }\nonumber \\ & =\frac {x}{\sqrt {x}}-x+\sqrt {x}\nonumber \\ & =\frac {x\sqrt {x}}{x}-x+\sqrt {x}\nonumber \\ & =2\sqrt {x}-x \tag {5}\end{align}
And substituting \(p_{2}\) into (1) gives
\begin{align} y_{1} & =xp_{1}+\frac {1}{1+p_{1}}\nonumber \\ & =x\left ( -\frac {1}{\sqrt {x}}-1\right ) +\frac {1}{1+\left ( -\frac {1}{\sqrt {x}}-1\right ) }\nonumber \\ & =-\frac {x}{\sqrt {x}}-x-\sqrt {x}\nonumber \\ & =\frac {-x\sqrt {x}}{x}-x-\sqrt {x}\nonumber \\ & =-2\sqrt {x}-x \tag {6}\end{align}
There are 3 solutions given in (4,5,6). One is general and two are singular.
2.2.15 Example 15
\[ xyy^{\prime }=y^{2}+x\sqrt {4x^{2}+y^{2}}\]
Solving for \(y\) gives
\begin{align*} {\normalsize y} & {\normalsize =}\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) {\normalsize x}\\ y & =xf+g \end{align*}
Where \(f=\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) \) and \(g=0\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative of the above w.r.t. \(x\)
gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+xf^{\prime }\frac {dp}{dx}\\ p-f & =xf^{\prime }\frac {dp}{dx}\end{align*}
Using values for \(f\ \ \)the above simplifies to
\begin{equation} p-\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) =\left ( x\frac {d}{dp}\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) \right ) \frac {dp}{dx} \tag {2A}\end{equation}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which
results in \(p=\operatorname {RootOf}\left ( \_z^{4}-4+\left ( p^{2}-1\right ) \_z^{2}-2\_z^{3}p\right ) \). Substituting this in (1) does not generate any real solutions (only 2 complex
ones) hence will not be used.
The general solution is found by finding \(p\) from (2A). Since (2A) is not linear in \(p\), then
inversion is needed. Writing (2A) as
\begin{align*} \frac {dx}{dp} & =\frac {xf}{p-f}\\ \frac {1}{x}dx & =\frac {f}{p-f}dp \end{align*}
Due to complexity of result, one now needs to obtain explicit result for \(\operatorname {RootOf}\) which makes the
computation very complicated. So this is not practical to solve by hand. Will stop here. It is
much easier to solve this ode as a homogeneous ode instead which gives the solution
as
\[ -\frac {\sqrt {4x^{2}+y^{2}}}{x}+\ln \left ( x\right ) =c_{1}\]
2.2.16 Example 16
\[ \ln \left ( \cos y^{\prime }\right ) +y^{\prime }\tan y^{\prime }=y \]
Solving for \(y\) gives
\begin{align} y & =\ln \left ( \cos p\right ) +p\tan p\tag {1}\\ y & =xf+g\nonumber \\ & =g \tag {1A}\end{align}
Where \(f=0\) and \(g\left ( p\right ) =\ln \left ( \cos p\right ) +p\tan p\). Important note: This ode has \(f=0\) which is strictly speaking is not of the form \(y=xf\left ( p\right ) +g\left ( p\right ) \).
But Maple says this is dAlembert. This is why it is included. I should make special case
dAlmbert classification to handle this special case.
Taking derivative of (1A) w.r.t. \(x\) gives
\begin{align} p & =\frac {dg}{dp}\frac {dp}{dx}\nonumber \\ p & =\left ( -\frac {\sin p}{\cos p}+\tan p+p\left ( 1+\tan ^{2}p\right ) \right ) \frac {dp}{dx}\nonumber \\ p & =\left ( -\tan p+\tan p+p\left ( 1+\tan ^{2}p\right ) \right ) \frac {dp}{dx}\nonumber \\ p & =p\left ( 1+\tan ^{2}p\right ) \frac {dp}{dx}\nonumber \\ 1 & =\left ( 1+\tan ^{2}p\right ) \frac {dp}{dx}\end{align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which does not result in solution.
The general solution is found by finding \(p\) from (2). Since (2) is not linear in \(p\), then inversion
is needed. Writing (1) as
\begin{align*} \frac {dx}{dp} & =1+\tan ^{2}p\\ dx & =\left ( 1+\tan ^{2}p\right ) dp \end{align*}
Integrating gives
\begin{align*} x & =\tan p+c\\ p & =\arctan \left ( x-c\right ) \end{align*}
Substituting the above in (1) gives the solution
\begin{align*} y & =\ln \left ( \cos p\right ) +p\tan p\\ & =\ln \left ( \cos \left ( \arctan \left ( x-c\right ) \right ) \right ) +\left ( \arctan \left ( x-c\right ) \right ) \tan \left ( \arctan \left ( x-c\right ) \right ) \\ & =\ln \left ( \cos \left ( \arctan \left ( x-c\right ) \right ) \right ) +\left ( x-c\right ) \arctan \left ( x-c\right ) \end{align*}
This ode also have solution \(y=0\).
2.2.17 Example 17
\[ x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x=0 \]
Solving for \(y\) gives
\begin{align} y & =x\left ( \frac {1}{2}y^{\prime }+2\frac {1}{y^{\prime }}\right ) \tag {1}\\ & =x\left ( \frac {1}{2}p+2\frac {1}{p}\right ) \nonumber \\ y & =xf\nonumber \end{align}
where \(f=\frac {1}{2}p+2\frac {1}{p},g=0\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative and simplifying gives
\begin{align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) \\ p-f & =xf^{\prime }\frac {dp}{dx}\end{align*}
Using values for \(f,g\) the above simplifies to
\begin{align} p-\frac {1}{2}p-2\frac {1}{p} & =x\left ( \frac {1}{2}-\frac {2}{p^{2}}\right ) \frac {dp}{dx}\nonumber \\ \frac {1}{2}p-\frac {2}{p} & =x\left ( \frac {1}{2}-\frac {2}{p^{2}}\right ) \frac {dp}{dx} \tag {2A}\end{align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(\frac {1}{2}p-\frac {2}{p}=0\) or \(\frac {1}{2}p^{2}-2=0\) or \(p^{2}=4\) or \(p=\pm 2\). Hence \(y=\pm 2x\) are the
singular solutions.
The general solution is when \(\frac {dp}{dx}\neq 0\) in (2A). Since (2A) is nonlinear, inversion is needed. General
solution can be shown to be
\begin{equation} y=-\frac {1}{2}\left ( -\frac {x^{2}}{c_{1}^{2}}-4\right ) c_{1} \tag {3}\end{equation}
Will now show a more general method to find singular solution
that works for any first order ode. This requires finding the general solution above first. Let
the general solution be
\begin{align*} \Phi \left ( x,y,c\right ) & =0\\ & =y+\frac {1}{2}\left ( -\frac {x^{2}}{c_{1}^{2}}-4\right ) c_{1}\end{align*}
The ode is
\begin{align*} F\left ( x,y,y^{\prime }\right ) & =0\\ & =x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x \end{align*}
First we find the p-discriminant curve. This is found by eliminating \(y^{\prime }\) from
\begin{align*} F & =0\\ \frac {\partial F}{\partial y^{\prime }} & =0 \end{align*}
Or
\begin{align*} x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x & =0\\ 2xy^{\prime }-2y & =0 \end{align*}
Second equation gives \(y^{\prime }=\frac {y}{x}\). Substituting into first equation gives \(x\left ( \frac {y}{x}\right ) ^{2}-2y\left ( \frac {y}{x}\right ) +4x=0\) or \(\frac {y^{2}}{x}-2\frac {y^{2}}{x}+4x=0\) or \(y=\pm 2x\). These are the
candidate singular solutions
\[ y_{s}=\pm 2x \]
Next, we verify these satisfy the ode itself. We see both do.
Next we have to check that for an arbitrary point \(x_{0}\) the following two equations are
satisfied
\begin{align*} y_{g}\left ( x_{0}\right ) & =y_{s}\left ( x_{0}\right ) \\ y_{g}^{\prime }\left ( x_{0}\right ) & =y_{s}^{\prime }\left ( x_{0}\right ) \end{align*}
Where \(y_{g}\left ( x\right ) \) is the general solution obtained above in (3). Starting with \(y_{s}=2x\) the above two equations
now become
\begin{align*} -\frac {1}{2}\left ( -\frac {x_{0}^{2}}{c_{1}^{2}}-4\right ) c_{1} & =2x_{0}\\ -\frac {1}{2}\left ( -\frac {2x_{0}}{c_{1}^{2}}\right ) c_{1} & =2 \end{align*}
Or
\begin{align*} \frac {x_{0}^{2}}{2c_{1}}+2c_{1} & =2x_{0}\\ \frac {x_{0}}{c_{1}} & =2 \end{align*}
Second equation gives \(c_{1}=\frac {x_{0}}{2}\). Using this in first equation gives
\begin{align*} \frac {x_{0}^{2}}{2\frac {x_{0}}{2}}+2\left ( \frac {x_{0}}{2}\right ) & =2x_{0}\\ x_{0}+x_{0} & =2x_{0}\\ 2x_{0} & =2x_{0}\end{align*}
Which shows it is satisfied. Hence this shows that \(y_{s}=2x\) is indeed a singular solution.
Now we have to do the same for second \(y_{s}=-2x\). Hence the steps of this method are the
following
- Find \(y_{s}\) using p-discriminant method by eliminating \(y^{\prime }\) from \(F=0\) and \(\frac {\partial F}{\partial y^{\prime }}=0\).
- Verify that each \(y_{s}\) found satisfies the ode.
- Find general solution to the ode \(y_{g}\left ( x\right ) \).
- Verify that the two equations \(y_{g}\left ( x_{0}\right ) =y_{s}\left ( x_{0}\right ) \) and \(y_{g}^{\prime }\left ( x_{0}\right ) =y_{s}^{\prime }\left ( x_{0}\right ) \) are satisfied at an arbitrary point \(x_{0}\). If so, then
\(y_{s}\) is singular solution. (envelope of the family of curves of the general solution).
2.2.18 Example 18
\[ x-yy^{\prime }=a\left ( y^{\prime }\right ) ^{2}\]
Solving for \(y\) gives
\begin{align} -yp & =-x+ap^{2}\nonumber \\ -y & =-\frac {x}{p}+ap\nonumber \\ y & =\frac {x}{p}-ap\tag {1}\\ y & =xf\left ( p\right ) +g\left ( p\right ) \nonumber \end{align}
Where \(f=\frac {1}{p},g=-ap\). Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative and simplifying gives
\begin{align*} p & =\frac {d}{dx}\left ( xf\left ( p\right ) +g\left ( p\right ) \right ) \\ & =f\left ( p\right ) +xf^{\prime }\left ( p\right ) \frac {dp}{dx}+g^{\prime }\left ( p\right ) \frac {dp}{dx}\end{align*}
But \(f\left ( p\right ) =\frac {1}{p},f^{\prime }\left ( p\right ) =\frac {-1}{p^{2}},g^{\prime }\left ( p\right ) =-a\) and the above becomes
\begin{align} p & =\frac {1}{p}-\frac {x}{p^{2}}\frac {dp}{dx}-a\frac {dp}{dx}\nonumber \\ p-\frac {1}{p} & =\left ( -\frac {x}{p^{2}}-a\right ) \frac {dp}{dx} \tag {2}\end{align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=\pm 1\). Hence \(y^{\prime }=\pm 1\) or \(y=\pm x\) but these do not
satisfy the ode, hence no singular solutions exist.
The general solution is when \(\frac {dp}{dx}\neq 0\) in (2). This gives the ode
\begin{align*} \frac {dp}{dx} & =\frac {p-\frac {1}{p}}{-\frac {x}{p^{2}}-a}\\ & =\frac {p-p^{3}}{ap^{2}+x}\end{align*}
But this is non-linear. Hence inversion is needed. This becomes
\[ \frac {dx}{dp}=\frac {-x\left ( p\right ) -ap^{2}}{p^{3}-p}\]
Which is now linear in \(x\left ( p\right ) \).
The solution is
\begin{equation} x=\frac {-pa\sqrt {\left ( p-1\right ) \left ( p+1\right ) }\ln \left ( p+\sqrt {p^{2}-1}\right ) }{\left ( p-1\right ) \left ( p+1\right ) }+\frac {pc_{1}}{\sqrt {p-1}\sqrt {p+1}} \tag {3}\end{equation}
From (1) \(y=\frac {x}{p}-ap\), hence
\begin{align*} p_{1} & =\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\\ p_{2} & =-\frac {1}{2}\frac {y+\sqrt {4ax+y^{2}}}{a}\end{align*}
Plugging \(p_{1}\) into (3) gives one solution and Plugging \(p_{2}\) into (3) gives the second solution.
2.2.19 Example 19
\[ y=xf\left ( p\right ) +g\left ( p\right ) \]
This problem is meant to show what to do when we are unable to solve explicitly for \(x\left ( p\right ) \) when
doing inversion. Taking derivative the above becomes
\begin{align*} p & =\frac {d}{dx}\left ( xf\left ( p\right ) +g\left ( p\right ) \right ) \\ & =f\left ( p\right ) +xf^{\prime }\left ( p\right ) \frac {dp}{dx}+g^{\prime }\left ( p\right ) \frac {dp}{dx}\\ p-f\left ( p\right ) & =\left ( xf^{\prime }\left ( p\right ) +g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ \frac {dp}{dx} & =\frac {p-f\left ( p\right ) }{\left ( xf^{\prime }\left ( p\right ) +g^{\prime }\left ( p\right ) \right ) }\end{align*}
Inversion is needed. Hence gives
\begin{align*} \frac {dx\left ( p\right ) }{dp} & =\frac {\left ( x\left ( p\right ) f^{\prime }\left ( p\right ) +g^{\prime }\left ( p\right ) \right ) }{p-f\left ( p\right ) }\\ \frac {dx}{dp} & =\frac {xf^{\prime }}{p-f}+\frac {g^{\prime }}{p-f}\end{align*}
This is now linear in \(x\).
\[ \frac {dx}{dp}-\frac {xf^{\prime }}{p-f}=\frac {g^{\prime }}{p-f}\]
Integrating factor is \(\mu =e^{\int \frac {f^{\prime }\left ( p\right ) }{p-f}dp}\). Hence the above becomes
\begin{align} \frac {d}{dp}\left ( x\mu \right ) & =\mu \frac {g^{\prime }}{p-f}\nonumber \\ x\mu & =\int \mu \frac {g^{\prime }}{p-f}dp+c_{1}\nonumber \\ x & =\frac {1}{\mu }\int \mu \frac {g^{\prime }}{p-f}dp+c_{1}\mu \tag {1}\end{align}
Now we solve for \(p\) from \(y=xf\left ( p\right ) +g\left ( p\right ) \) and plug-in the result into the above. To show how this work, lets
apply the earlier problem to the above which was to solve \(x-yy^{\prime }=a\left ( y^{\prime }\right ) ^{2}\). From that problem we found
that
\begin{align*} p_{1} & =\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\\ p_{2} & =-\frac {1}{2}\frac {y+\sqrt {4ax+y^{2}}}{a}\end{align*}
And we had \(f=\frac {1}{p},g=-ap\). Using these value we now find
\begin{align*} \mu & =e^{\int \frac {f^{\prime }\left ( p\right ) }{p-f}dp}\\ & =e^{\int \frac {-\frac {1}{p^{2}}}{p-\frac {1}{p}}dp}\\ & =\frac {p}{\sqrt {p^{2}-1}}\end{align*}
Hence
\begin{align*} x & =\frac {\sqrt {p^{2}-1}}{p}\int \frac {p}{\sqrt {p^{2}-1}}\frac {-a}{p-\frac {1}{p}}dp+c_{1}\frac {p}{\sqrt {p^{2}-1}}\\ & =-\frac {a\sqrt {p^{2}-1}}{p}\int \frac {p^{2}}{\left ( p^{2}-1\right ) ^{\frac {3}{2}}}dp+c_{1}\frac {p}{\sqrt {p^{2}-1}}\\ & =-\frac {a\sqrt {p^{2}-1}}{p}\left ( -\frac {p}{\sqrt {p^{2}-1}}+\ln \left ( p+\sqrt {p^{2}-1}\right ) \right ) +c_{1}\frac {p}{\sqrt {p^{2}-1}}\\ & =a-\frac {a\sqrt {p^{2}-1}}{p}\ln \left ( p+\sqrt {p^{2}-1}\right ) +c_{1}\frac {p}{\sqrt {p^{2}-1}}\end{align*}
Substituting each one of the above value for \(p\) in (2) gives the two solutions. For example,
using \(p_{1}=\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\) gives
\[ x=a-\frac {a\sqrt {\left ( \frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\right ) ^{2}-1}}{\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}}\ln \left ( \frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}+\sqrt {\left ( \frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\right ) ^{2}-1}\right ) +c_{1}\frac {\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}}{\sqrt {\left ( \frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\right ) ^{2}-1}}\]
And same for the other \(p_{2}\).
In the above example it was possible to evaluate the integrals in \(p\), then replace \(p\) by its
solution from the original ode. What if this was not possible? Let say we have
integral
\[ \int ap^{2}dp \]
And for some reason we are not able to the integration. In this case we
first replace the above with
\[ \int ^{p}a\tau ^{2}d\tau \]
And only now replace \(p\) with its solution as the upper
limit.
2.2.20 Example 20
\[ y^{\prime }=-\frac {x}{2}-1+\frac {1}{2}\sqrt {x^{2}+4x+4y}\]
Solving for \(y\) gives
\begin{align} y & =xp+\left ( 1+2p+p^{2}\right ) \tag {1}\\ y & =xf+g\nonumber \end{align}
Hence \(f=p,g=\left ( 1+2p+p^{2}\right ) \). Since \(f=p\) then this is Clairaut. Taking derivative of the above w.r.t. \(x\) gives
\begin{align*} y^{\prime } & =f+x\frac {df}{dp}\frac {dp}{dx}+\frac {dg}{dp}\frac {dp}{dx}\\ p & =f+\frac {dp}{dx}\left ( x\frac {df}{dp}+\frac {dg}{dp}\right ) \end{align*}
But \(\frac {df}{dp}=1,\frac {dg}{dp}=2+2p\). The above becomes
\[ p-f=\frac {dp}{dx}\left ( x+2+2p\right ) \]
But \(f=p\). The above simplifies to
\begin{equation} 0=\frac {dp}{dx}\left ( x+2+2p\right ) \tag {2}\end{equation}
The general solution is when \(\frac {dp}{dx}=0\).
Hence \(p=c_{1}\). Substituting this into (1) gives
\[ y=xc_{1}+\left ( 1+2c_{1}+c_{1}^{2}\right ) \]
The singular solution is when \(\frac {dp}{dx}\neq 0\) in (2) which
gives
\begin{align*} x+2+2p & =0\\ p & =\frac {-x-2}{2}\end{align*}
Substituting this in (1) gives
\begin{align*} y & =x\left ( \frac {-x-2}{2}\right ) +\left ( 1+2\left ( \frac {-x-2}{2}\right ) +\left ( \frac {-x-2}{2}\right ) ^{2}\right ) \\ & =-\frac {1}{4}x\left ( x+4\right ) \\ & =-\frac {1}{4}x^{2}-x \end{align*}
Checking this solution against the ode shows it is verifies the ode. Hence there are two
solutions, one general and one singular
\[ y=\left \{ \begin {array} [c]{c}xc_{1}+1+2c_{1}+c_{1}^{2}\\ -\frac {1}{4}x^{2}-x \end {array} \right . \]
2.2.21 Example 21
\[ \frac {y^{\prime }y}{1+\frac {1}{2}\sqrt {1+\left ( y^{\prime }\right ) ^{2}}}=-x \]
Let \(y^{\prime }=p\) and rearranging gives
\begin{align} py & =-x\left ( 1+\frac {1}{2}\sqrt {1+p^{2}}\right ) \nonumber \\ y & =-x\left ( \frac {1}{p}+\frac {1}{2p}\sqrt {1+p^{2}}\right ) \nonumber \\ & =-x\left ( \frac {2}{2p}+\frac {1}{2p}\sqrt {1+p^{2}}\right ) \nonumber \\ & =-x\left ( \frac {2+\sqrt {1+p^{2}}}{2p}\right ) \nonumber \\ & =xf+g \tag {1}\end{align}
Hence
\begin{align*} f & =-\frac {2+\sqrt {1+p^{2}}}{2p}\\ g & =0 \end{align*}
Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative of (1) w.r.t. \(x\) gives
\begin{align*} p & =\frac {d}{dx}\left ( xf\left ( p\right ) +g\left ( p\right ) \right ) \\ & =f\left ( p\right ) +xf^{\prime }\left ( p\right ) \frac {dp}{dx}+g^{\prime }\left ( p\right ) \frac {dp}{dx}\end{align*}
But \(f\left ( p\right ) =-\frac {2+\sqrt {1+p^{2}}}{2p},f^{\prime }\left ( p\right ) =\frac {-1}{p^{2}},g=0,g^{\prime }=0\) and the above becomes
\begin{align} p & =-\frac {2+\sqrt {1+p^{2}}}{2p}+x\left ( -\frac {1}{2\sqrt {1+p^{2}}}-\frac {-2-\sqrt {1+p^{2}}}{2p^{2}}\right ) \frac {dp}{dx}\nonumber \\ p+\frac {2+\sqrt {1+p^{2}}}{2p} & =x\left ( -\frac {1}{2\sqrt {1+p^{2}}}-\frac {-2-\sqrt {1+p^{2}}}{2p^{2}}\right ) \frac {dp}{dx} \tag {2}\end{align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p+\frac {2+\sqrt {1+p^{2}}}{2p}=0\). Hence \(p=\pm i\) or \(y^{\prime }=\pm i\) \(\ \) or \(y=\pm ix\). But these do not
satisfy the ode, hence no singular solutions exist.
The general solution is when \(\frac {dp}{dx}\neq 0\) in (2). This gives the ode
\begin{align*} \frac {dp}{dx} & =\frac {1}{x}\frac {\left ( p+\frac {2+\sqrt {1+p^{2}}}{2p}\right ) }{\left ( -\frac {1}{2\sqrt {1+p^{2}}}-\frac {-2-\sqrt {1+p^{2}}}{2p^{2}}\right ) }\\ & =\frac {1}{x}\left ( p^{3}+p\right ) \end{align*}
But this is non-linear in \(p\). Hence inversion is needed. This becomes
\begin{align*} \frac {dx}{dp} & =x\frac {\left ( -\frac {1}{2\sqrt {1+p^{2}}}-\frac {-2-\sqrt {1+p^{2}}}{2p^{2}}\right ) }{\left ( p+\frac {2+\sqrt {1+p^{2}}}{2p}\right ) }\\ \frac {dx}{dp} & =\frac {x}{p^{3}+p}\\ \frac {dx}{dp}-\frac {1}{p+p^{3}}x & =0 \end{align*}
Which is now linear in \(x\left ( p\right ) \). The solution is
\begin{equation} x=\frac {p}{\sqrt {1+p^{2}}}c_{1} \tag {3}\end{equation}
We now need to eliminate \(p\). We have two equations
to do that, (1) and (3). Here they are side by side
\begin{align} y & =-x\left ( \frac {2+\sqrt {1+p^{2}}}{2p}\right ) \tag {1}\\ x & =\frac {p}{\sqrt {1+p^{2}}}c_{1} \tag {3}\end{align}
We can either solve for \(p\) from (1) and plugin in the value found into (3). Or we can solve for \(p\)
from (3) and plugin the value found in (1). Using CAS we can just use the solve command.
For an example, using Maple it gives
eq1:=y=-x*( (2+sqrt(1+p^2))/(2*p));
eq2:=x=p/sqrt(1+p^2)*_C1
sol:=solve([eq1,eq2],[p,y],'allsolutions');
[[p = x*RootOf((c__1^2 - x^2)*_Z^2 - 1), y = -(RootOf((c__1^2 - x^2)*_Z^2 - 1)*c__1 + 2)/(2*RootOf((c__1^2 - x^2)*_Z^2 - 1))]]
Now we can use allvalues
map(X->allvalues(X),sol)
[[p = x*sqrt(1/(c__1^2 - x^2)), y = -(sqrt(1/(c__1^2 - x^2))*c__1 + 2)/(2*sqrt(1/(c__1^2 - x^2)))],
[p = -x*sqrt(1/(c__1^2 - x^2)), y = (-sqrt(1/(c__1^2 - x^2))*c__1 + 2)/(2*sqrt(1/(c__1^2 - x^2)))]]
Hence the solutions are
\begin{align*} y_{1} & =-\frac {\sqrt {\frac {1}{c_{1}^{2}-x^{2}}}c_{1}+2}{2\sqrt {\frac {1}{c_{1}^{2}-x^{2}}}}\\ y_{2} & =-\frac {-\sqrt {\frac {1}{c_{1}^{2}-x^{2}}}c_{1}+2}{2\sqrt {\frac {1}{c_{1}^{2}-x^{2}}}}\end{align*}
These are verified valid solutions to the ode (had to use assuming positive)
2.2.22 Example 22
\[ x\left ( y^{\prime }\right ) ^{3}=yy^{\prime }+1 \]
Let \(y^{\prime }=p\) and rearranging gives
\begin{align} xp^{3} & =yp+1\nonumber \\ y & =\frac {xp^{3}-1}{p}\nonumber \\ & =xp^{2}-\frac {1}{p}\nonumber \\ & =xf+g \tag {1}\end{align}
Hence
\begin{align*} f & =p^{2}\\ g & =-\frac {1}{p}\end{align*}
Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative of (1) w.r.t. \(x\) gives
\begin{align*} p & =\frac {d}{dx}\left ( xf\left ( p\right ) +g\left ( p\right ) \right ) \\ & =f\left ( p\right ) +xf^{\prime }\left ( p\right ) \frac {dp}{dx}+g^{\prime }\left ( p\right ) \frac {dp}{dx}\\ & =f\left ( p\right ) +\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}
But \(f\left ( p\right ) =p^{2},f^{\prime }\left ( p\right ) =2p,g=-\frac {1}{p},g^{\prime }=\frac {1}{p^{2}}\) and the above becomes
\begin{align} p & =p^{2}+\left ( 2xp+\frac {1}{p^{2}}\right ) \frac {dp}{dx}\nonumber \\ p-p^{2} & =\left ( 2xp+\frac {1}{p^{2}}\right ) \frac {dp}{dx} \tag {2}\end{align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p-p^{2}=0\). Hence \(p=0\) or \(p=1\). Substituting \(p=0\) in (1)
gives 1/0 error. Hence this is not valid solution. Substituting \(p=1\) in (1) gives \(y=x-1\) which verifies the
ode. Hence this is valid singular solution.
The general solution is when \(\frac {dp}{dx}\neq 0\) in (2). This gives the ode
\[ \frac {dp}{dx}=\frac {p^{3}\left ( 1-p\right ) }{2xp^{3}+1}\]
But this is non-linear in \(p\). Hence
inversion is needed. This becomes
\[ \frac {dx}{dp}=\frac {2xp^{3}+1}{p^{3}\left ( 1-p\right ) }\]
Which is now linear in \(x\left ( p\right ) \). The solution is
\begin{equation} x=\frac {2c_{1}p^{2}+2p-1}{2p^{2}\left ( p-1\right ) ^{2}} \tag {3}\end{equation}
We now need to
eliminate \(p\). We have two equations to do that, (1) and (3). Here they are side by
side
\begin{align} y & =xp^{2}-\frac {1}{p}\tag {1}\\ x & =\frac {2c_{1}p^{2}+2p-1}{2p^{2}\left ( p-1\right ) ^{2}} \tag {3}\end{align}
We can either solve for \(p\) from (1) and plugin in the value found into (3). Or we can solve for \(p\)
from (3) and plugin the value found in (1). Using CAS we can just use the solve command.
For an example, using Maple it gives
eq1:=y=x*p^2-1/p;
eq2:=x= (2*_C1*p^2+2*p-1)/(2*p^2*(p-1)^2);
solve({eq1,eq2},{y,p})
Whch gives
{p = RootOf(1 + 2*x*_Z^4 - 4*x*_Z^3 + (-2*c__1 + 2*x)*_Z^2 - 2*_Z),
y = (x*RootOf(1 + 2*x*_Z^4 - 4*x*_Z^3 + (-2*c__1 + 2*x)*_Z^2 - 2*_Z)^3 - 1)/RootOf(1 + 2*x*_Z^4 - 4*x*_Z^3 + (-2*c__1 + 2*x)*_Z^2 - 2*_Z)}
Hence the general solution is
\[ y=\frac {x\operatorname {RootOf}\left ( 1+2xZ^{4}-4xZ^{3}+\left ( -2c_{1}+2x\right ) Z^{2}-2Z\right ) ^{3}-1}{\operatorname {RootOf}\left ( 1+2xZ^{4}-4xZ^{3}+\left ( -2c_{1}+2x\right ) Z^{2}-2Z\right ) }\]
And the singular solution is
\[ y=x-1 \]
2.2.23 Example 23
\[ \left ( y^{\prime }\right ) ^{2}-2yy^{\prime }=2x \]
Let \(y^{\prime }=p\) and rearranging gives
\begin{align} p^{2}-2yp & =2x\nonumber \\ y & =\frac {p^{2}-2x}{2p}\nonumber \\ & =-x\frac {1}{p}+\frac {1}{2}p\nonumber \\ & =xf+g \tag {1}\end{align}
Hence
\begin{align*} f & =-\frac {1}{p}\\ g & =\frac {1}{2}p \end{align*}
Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative of (1) w.r.t. \(x\) gives
\begin{align*} p & =\frac {d}{dx}\left ( xf\left ( p\right ) +g\left ( p\right ) \right ) \\ & =f\left ( p\right ) +xf^{\prime }\left ( p\right ) \frac {dp}{dx}+g^{\prime }\left ( p\right ) \frac {dp}{dx}\\ & =f\left ( p\right ) +\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\end{align*}
But \(f\left ( p\right ) =-\frac {1}{p},f^{\prime }\left ( p\right ) =\frac {1}{p^{2}},g=\frac {1}{2}p,g^{\prime }=\frac {1}{2}\) and the above becomes
\begin{align} p & =-\frac {1}{p}+\left ( \frac {x}{p^{2}}+\frac {1}{2}\right ) \frac {dp}{dx}\nonumber \\ p+\frac {1}{p} & =\left ( \frac {x}{p^{2}}+\frac {1}{2}\right ) \frac {dp}{dx} \tag {2}\end{align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p^{2}+1=0\). Hence \(p=\pm i\) But these do not verify
the ode. Hence no singular solutions exist.
The general solution is when \(\frac {dp}{dx}\neq 0\) in (2). This gives the ode
\[ \frac {dp}{dx}=\frac {\left ( p^{2}+1\right ) 2p}{2x+p^{2}}\]
But this is non-linear in \(p\). Hence
inversion is needed. This becomes
\[ \frac {dx}{dp}=\frac {2x+p^{2}}{\left ( p^{2}+1\right ) 2p}\]
Which is now linear in \(x\left ( p\right ) \). The solution is
\begin{equation} x=\frac {\left ( \frac {1}{2}\operatorname {arcsinh}\left ( p\right ) +c_{1}\right ) p}{\sqrt {p^{2}-1}} \tag {3}\end{equation}
We now need to
eliminate \(p\). We have two equations to do that, (1) and (3). Here they are side by
side
\begin{align} y & =-x\frac {1}{p}+\frac {1}{2}p\tag {1}\\ x & =\frac {\left ( \frac {1}{2}\operatorname {arcsinh}\left ( p\right ) +c_{1}\right ) p}{\sqrt {p^{2}-1}} \tag {3}\end{align}
We can either solve for \(p\) from (1) and plugin in the value found into (3). Or we can solve for \(p\)
from (3) and plugin the value found in (1). In this case it is easier to solve for \(p\) from (1)
which gives
\begin{align*} p_{1} & =y+\sqrt {2x+y^{2}}\\ p_{2} & =y-\sqrt {2x+y^{2}}\end{align*}
Substituting each of these into (3) gives these two general solutions
\begin{align*} x & =\frac {\left ( \frac {1}{2}\operatorname {arcsinh}\left ( y+\sqrt {2x+y^{2}}\right ) +c_{1}\right ) \left ( y+\sqrt {2x+y^{2}}\right ) }{\sqrt {\left ( y+\sqrt {2x+y^{2}}\right ) ^{2}-1}}\\ x & =\frac {\left ( \frac {1}{2}\operatorname {arcsinh}\left ( y-\sqrt {2x+y^{2}}\right ) +c_{1}\right ) \left ( y-\sqrt {2x+y^{2}}\right ) }{\sqrt {\left ( y-\sqrt {2x+y^{2}}\right ) ^{2}-1}}\end{align*}
2.2.24 Example 24
\[ xy^{\prime }-y=\sqrt {x^{2}-y^{2}}\]
Let \(y^{\prime }=p\) and rearranging gives
\[ xp-y=\sqrt {x^{2}-y^{2}}\]
Solving for \(y\) gives two solutions
\begin{align} y & =x\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) \tag {1}\\ y & =x\left ( \frac {p}{2}-\frac {1}{2}\sqrt {2-p^{2}}\right ) \nonumber \end{align}
We will here solve the first one above. The second one will have similar solution. Comparing
the above to \(y=xf\left ( p\right ) +g\left ( p\right ) \) shows that
\begin{align} f & =\frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\tag {2}\\ g & =0\nonumber \end{align}
Since \(f\left ( p\right ) \neq p\) then this is d’Almbert ode. Taking derivative of (2) w.r.t. \(x\) gives
\begin{align} p & =\frac {d}{dx}\left ( xf\left ( p\right ) \right ) \nonumber \\ & =f\left ( p\right ) +xf^{\prime }\left ( p\right ) \frac {dp}{dx}\nonumber \\ & =\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) +x\left ( \frac {1}{2}-\frac {p}{2\sqrt {2-p^{2}}}\right ) \frac {dp}{dx}\nonumber \\ p-\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) & =x\left ( \frac {1}{2}-\frac {p}{2\sqrt {2-p^{2}}}\right ) \frac {dp}{dx}\tag {3}\end{align}
Singular solution is when \(\frac {dp}{dx}=0\) which results in
\begin{align*} p-\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) & =0\\ \frac {p}{2}-\frac {1}{2}\sqrt {2-p^{2}} & =0 \end{align*}
Hence \(p=1\). Substituting this in (2) gives singular solution
\begin{align*} y & =x\left ( \frac {1}{2}+\frac {1}{2}\sqrt {2-1}\right ) \\ & =x \end{align*}
To find general solution, we need to solve (3) for \(p\). EQ (3) becomes
\begin{align*} \frac {dp}{dx} & =\frac {\frac {p}{2}-\frac {1}{2}\sqrt {2-p^{2}}}{\frac {x}{2}-\frac {xp}{2\sqrt {2-p^{2}}}}\\ & =-\frac {1}{x}\sqrt {2-p^{2}}\end{align*}
This is separable ode.
\begin{align*} \frac {-dp}{\sqrt {2-p^{2}}} & =\frac {1}{x}dx\\ -\arcsin \left ( \frac {\sqrt {2}}{2}p\right ) & =\ln x+c_{1}\end{align*}
Substituting this into (1) gives
\begin{align*} y & =x\left ( \frac {p}{2}+\frac {1}{2}\sqrt {2-p^{2}}\right ) \\ & =x\left ( \frac {-\frac {2}{\sqrt {2}}\sin \left ( \ln x+c_{1}\right ) }{2}+\frac {1}{2}\sqrt {2-\left ( -\frac {2}{\sqrt {2}}\sin \left ( \ln x+c_{1}\right ) \right ) ^{2}}\right ) \\ & =x\left ( \frac {-\sin \left ( \ln x+c_{1}\right ) }{\sqrt {2}}+\frac {1}{2}\sqrt {2-2\sin ^{2}\left ( \ln x+c_{1}\right ) }\right ) \end{align*}
This ode is an example where \(y\) does not appear explicitly in the ode so not possible to
directly solve for \(y\). It is given here to show possible problems with this method.
\begin{equation} y^{\prime }=\sqrt {1+x+y} \tag {1A}\end{equation}
This ode is
squared to first solve for \(y\) which gives
\begin{equation} \left ( y^{\prime }\right ) ^{2}=1+x+y \tag {2A}\end{equation}
However, here care is needed. To get back to original
ode (1A) then (2A) means two possible equations
\[ y^{\prime }=\pm \sqrt {1+x+y}\]
Hence the solutions obtained using (2A)
can be the solution to one of these
\begin{align} y^{\prime } & =+\sqrt {1+x+y}\tag {B1}\\ y^{\prime } & =-\sqrt {1+x+y} \tag {B2}\end{align}
Therefore the solution obtained by squaring both sides of (1A), which is done in order to
solve for \(y\), must be checked to see if it satisfies the original ode, else it will be extraneous
solution resulting from squaring both sides of the ode.
Starting from (2A), in normal form (by replacing \(y^{\prime }\) with \(p\)) it becomes
\begin{align} y & =-x-1+p^{2}\tag {1}\\ & =xf+g\nonumber \end{align}
Where \(f=-1,g=-1+p^{2}\). Taking derivative w.r.t. \(x\) gives
\begin{align} p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\nonumber \\ p+1 & =2p\frac {dp}{dx} \tag {2}\end{align}
Since \(\frac {\partial \phi }{\partial x}=-1\neq p\) then this is d’Alembert ode. The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in
\(p=-1\). Substituting this in (1) gives the singular solution
\begin{equation} y\left ( x\right ) =-x \tag {3}\end{equation}
But this solution does not satisfy the
ode, hence it is extraneous. The general solution is found by finding \(p\) from (2). Since (2) is
nonlinear, then it is inverted which gives
\begin{align*} \frac {p+1}{2p} & =\frac {dp}{dx}\\ \frac {dx}{dp} & =\frac {2p}{p+1}\end{align*}
Which is linear in \(x\). Solving gives
\begin{equation} x=2p-2\ln \left ( p+1\right ) +c_{1} \tag {4}\end{equation}
Instead of inverting this to find \(p\) in terms of \(x\), \(p\) is found from
(1) which gives
\begin{align*} y+x+1 & =p^{2}\\ p & =\pm \sqrt {y+x+1}\end{align*}
Substituting these solutions in (4) gives implicit solutions as
\begin{align*} x & =2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_{1}\\ x & =-2\sqrt {y+x+1}-2\ln \left ( 1-\sqrt {y+x+1}\right ) +c_{1}\end{align*}
But only the first one above satisfies the ode. The second is extraneous. Therefore the final
solution is
\[ x=2\sqrt {y+x+1}-2\ln \left ( 1+\sqrt {y+x+1}\right ) +c_{1}\]
And no singular solutions exist. If instead of doing the above, \(p\) was found from (4)
using inversion, then it will be
\[ p=-\operatorname *{LambertW}\left ( -c_{1}e^{\frac {-x}{2}-1}\right ) -1 \]
Substituting this in (1) gives
\[ y=-x-1+\left ( -\operatorname *{LambertW}\left ( -c_{1}e^{\frac {-x}{2}-1}\right ) -1\right ) ^{2}\]
But this general solution
does not satisfy the original ode. In general, it is best to avoid squaring both
side of the ode in order to solve for \(y\) as this can generate extraneous solutions.
Only use this method if the original ode is already given in the form where \(y\) shows
explicitly.