2.1.13 Problem 13
Internal
problem
ID
[9084]
Book
:
Second
order
enumerated
odes
Section
:
section
1
Problem
number
:
13
Date
solved
:
Monday, January 27, 2025 at 05:32:17 PM
CAS
classification
:
[[_2nd_order, _missing_x]]
Solve
\begin{align*} {y^{\prime \prime }}^{2}+y^{\prime }&=0 \end{align*}
Solved as second order missing x ode
Time used: 2.523 (sec)
This is missing independent variable second order ode. Solved by reduction of order by using
substitution which makes the dependent variable \(y\) an independent variable. Using
\begin{align*} y' &= p \end{align*}
Then
\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}
Hence the ode becomes
\begin{align*} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 0 \end{align*}
Which is now solved as first order ode for \(p(y)\).
Factoring the ode gives these factors
\begin{align*}
\tag{1} p &= 0 \\
\tag{2} {p^{\prime }}^{2} p+1 &= 0 \\
\end{align*}
Now each of the above equations is solved in
turn.
Solving equation (1)
Solving for \(p\) from
\begin{align*} p = 0 \end{align*}
Solving gives \(p = 0\)
Solving equation (2)
Let \(p=p^{\prime }\) the ode becomes
\begin{align*} p^{2} p +1 = 0 \end{align*}
Solving for \(p\) from the above results in
\begin{align*}
\tag{1} p &= -\frac {1}{p^{2}} \\
\end{align*}
This has the form
\begin{align*} p=yf(p)+g(p)\tag {*} \end{align*}
Where \(f,g\) are functions of \(p=p'(y)\). The above ode is dAlembert ode which is now solved.
Taking derivative of (*) w.r.t. \(y\) gives
\begin{align*} p &= f+(y f'+g') \frac {dp}{dy}\\ p-f &= (y f'+g') \frac {dp}{dy}\tag {2} \end{align*}
Comparing the form \(p=y f + g\) to (1A) shows that
\begin{align*} f &= 0\\ g &= -\frac {1}{p^{2}} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p = \frac {2 p^{\prime }\left (y \right )}{p^{3}}
\end{equation}
The singular solution is found by setting \(\frac {dp}{dy}=0\) in the above which gives
\begin{align*} p = 0 \end{align*}
No valid singular solutions found.
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}}\neq 0\). From eq. (2A). This results in
\begin{equation}
\tag{3} p^{\prime }\left (y \right ) = \frac {p \left (y \right )^{4}}{2}
\end{equation}
This ODE is now solved
for \(p \left (y \right )\). No inversion is needed.
Integrating gives
\begin{align*} \int \frac {2}{p^{4}}d p &= dy\\ -\frac {2}{3 p^{3}}&= y +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} \frac {p^{4}}{2}&= 0 \end{align*}
for \(p \left (y \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} p \left (y \right ) = 0 \end{align*}
Solving for \(p \left (y \right )\) gives
\begin{align*}
p \left (y \right ) &= 0 \\
p \left (y \right ) &= \frac {\left (-18 \left (y +c_1 \right )^{2}\right )^{{1}/{3}}}{3 c_1 +3 y} \\
p \left (y \right ) &= -\frac {\left (-18 \left (y +c_1 \right )^{2}\right )^{{1}/{3}}}{6 \left (y +c_1 \right )}-\frac {i \sqrt {3}\, \left (-18 \left (y +c_1 \right )^{2}\right )^{{1}/{3}}}{6 \left (y +c_1 \right )} \\
p \left (y \right ) &= -\frac {\left (-18 \left (y +c_1 \right )^{2}\right )^{{1}/{3}}}{6 \left (y +c_1 \right )}+\frac {i \sqrt {3}\, \left (-18 \left (y +c_1 \right )^{2}\right )^{{1}/{3}}}{6 y +6 c_1} \\
\end{align*}
Substituing the above solution for \(p\) in (2A) gives
\begin{align*} p = -\frac {9 \left (y +c_1 \right )^{2}}{\left (-18 \left (y +c_1 \right )^{2}\right )^{{2}/{3}}}\\ p = -\frac {18 \left (y +c_1 \right )^{2}}{\left (-18 \left (y +c_1 \right )^{2}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )}\\ p = \frac {18 \left (y +c_1 \right )^{2}}{\left (-18 \left (y +c_1 \right )^{2}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )}\\ \end{align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = 0 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {0\, dx} + c_2 \\ y &= c_2 \end{align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = -\frac {9 \left (y+c_1 \right )^{2}}{\left (-18 \left (y+c_1 \right )^{2}\right )^{{2}/{3}}} \end{align*}
Integrating gives
\begin{align*} \int -\frac {\left (-18 \left (y +c_1 \right )^{2}\right )^{{2}/{3}}}{9 \left (y +c_1 \right )^{2}}d y &= dx\\ -\frac {18^{{2}/{3}} \left (-\left (y +c_1 \right )^{2}\right )^{{2}/{3}}}{3 c_1 +3 y}&= x +c_3 \end{align*}
For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = \frac {18 \left (y+c_1 \right )^{2}}{\left (-18 \left (y+c_1 \right )^{2}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )} \end{align*}
Integrating gives
\begin{align*} \int \frac {\left (-18 \left (y +c_1 \right )^{2}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )}{18 \left (y +c_1 \right )^{2}}d y &= dx\\ \frac {18^{{2}/{3}} \left (-\left (y +c_1 \right )^{2}\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )}{6 y +6 c_1}&= x +c_4 \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= -\frac {1}{12} c_4^{3}-\frac {1}{4} c_4^{2} x -\frac {1}{4} c_4 \,x^{2}-\frac {1}{12} x^{3}-c_1 \\
\end{align*}
For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order
ode to solve which is
\begin{align*} y^{\prime } = -\frac {18 \left (y+c_1 \right )^{2}}{\left (-18 \left (y+c_1 \right )^{2}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )} \end{align*}
Integrating gives
\begin{align*} \int -\frac {\left (-18 \left (y +c_1 \right )^{2}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )}{18 \left (y +c_1 \right )^{2}}d y &= dx\\ -\frac {18^{{2}/{3}} \left (-\left (y +c_1 \right )^{2}\right )^{{2}/{3}} \left (i \sqrt {3}-1\right )}{6 y +6 c_1}&= x +c_5 \end{align*}
Will add steps showing solving for IC soon.
Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not
verify)
\begin{align*} y&=c_2\\ y&=-\frac {1}{12} c_3^{3}-\frac {1}{4} x \,c_3^{2}-\frac {1}{4} c_3 \,x^{2}-\frac {1}{12} x^{3}-c_1\\ y&=-\frac {1}{12} c_4^{3}-\frac {1}{4} c_4^{2} x -\frac {1}{4} c_4 \,x^{2}-\frac {1}{12} x^{3}-c_1\\ y&=-\frac {1}{12} c_5^{3}-\frac {1}{4} c_5^{2} x -\frac {1}{4} c_5 \,x^{2}-\frac {1}{12} x^{3}-c_1 \end{align*}
Summary of solutions found
\begin{align*}
y &= c_2 \\
y &= -\frac {1}{12} c_3^{3}-\frac {1}{4} x \,c_3^{2}-\frac {1}{4} c_3 \,x^{2}-\frac {1}{12} x^{3}-c_1 \\
y &= -\frac {1}{12} c_4^{3}-\frac {1}{4} c_4^{2} x -\frac {1}{4} c_4 \,x^{2}-\frac {1}{12} x^{3}-c_1 \\
y &= -\frac {1}{12} c_5^{3}-\frac {1}{4} c_5^{2} x -\frac {1}{4} c_5 \,x^{2}-\frac {1}{12} x^{3}-c_1 \\
\end{align*}
Solved as second order missing y ode
Time used: 0.494 (sec)
This is second order ode with missing dependent variable \(y\). Let
\begin{align*} u(x) &= y^{\prime } \end{align*}
Then
\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}
Hence the ode becomes
\begin{align*} {u^{\prime }\left (x \right )}^{2}+u \left (x \right ) = 0 \end{align*}
Which is now solved for \(u(x)\) as first order ode.
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} u^{\prime }\left (x \right )&=\sqrt {-u \left (x \right )} \\
\tag{2} u^{\prime }\left (x \right )&=-\sqrt {-u \left (x \right )} \\
\end{align*}
Now each of the above is solved
separately.
Solving Eq. (1)
Integrating gives
\begin{align*} \int \frac {1}{\sqrt {-u}}d u &= dx\\ -2 \sqrt {-u}&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} \sqrt {-u}&= 0 \end{align*}
for \(u \left (x \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} u \left (x \right ) = 0 \end{align*}
Solving Eq. (2)
Integrating gives
\begin{align*} \int -\frac {1}{\sqrt {-u}}d u &= dx\\ 2 \sqrt {-u}&= x +c_2 \end{align*}
Singular solutions are found by solving
\begin{align*} -\sqrt {-u}&= 0 \end{align*}
for \(u \left (x \right )\). This is because we had to divide by this in the above step. This gives the following
singular solution(s), which also have to satisfy the given ODE.
\begin{align*} u \left (x \right ) = 0 \end{align*}
In summary, these are the solution found for \(u(x)\)
\begin{align*}
-2 \sqrt {-u \left (x \right )} &= x +c_1 \\
2 \sqrt {-u \left (x \right )} &= x +c_2 \\
u \left (x \right ) &= 0 \\
\end{align*}
For solution \(-2 \sqrt {-u \left (x \right )} = x +c_1\), since \(u=y^{\prime }\) then we now have a new
first order ode to solve which is
\begin{align*} -2 \sqrt {-y^{\prime }} = x +c_1 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {-\frac {1}{4} x^{2}-\frac {1}{2} c_1 x -\frac {1}{4} c_1^{2}\, dx}\\ y &= -\frac {\left (x +c_1 \right )^{3}}{12} + c_3 \end{align*}
For solution \(2 \sqrt {-u \left (x \right )} = x +c_2\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} 2 \sqrt {-y^{\prime }} = x +c_2 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {-\frac {1}{4} x^{2}-\frac {1}{2} c_2 x -\frac {1}{4} c_2^{2}\, dx}\\ y &= -\frac {\left (x +c_2 \right )^{3}}{12} + c_4 \end{align*}
For solution \(u \left (x \right ) = 0\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = 0 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).
\begin{align*} \int {dy} &= \int {0\, dx} + c_5 \\ y &= c_5 \end{align*}
In summary, these are the solution found for \((y)\)
\begin{align*}
y &= -\frac {\left (x +c_1 \right )^{3}}{12}+c_3 \\
y &= -\frac {\left (x +c_2 \right )^{3}}{12}+c_4 \\
y &= c_5 \\
\end{align*}
Will add steps showing solving for IC
soon.
Summary of solutions found
\begin{align*}
y &= c_5 \\
y &= -\frac {\left (x +c_1 \right )^{3}}{12}+c_3 \\
y &= -\frac {\left (x +c_2 \right )^{3}}{12}+c_4 \\
\end{align*}
Maple step by step solution
Maple trace
`Methods for second order ODEs:
*** Sublevel 2 ***
Methods for second order ODEs:
Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE.
*** Sublevel 3 ***
Methods for second order ODEs:
--- Trying classification methods ---
trying 2nd order Liouville
trying 2nd order WeierstrassP
trying 2nd order JacobiSN
differential order: 2; trying a linearization to 3rd order
trying 2nd order ODE linearizable_by_differentiation
-> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x)+1/2, y(x)` *** Sublevel 4 ***
Methods for third order ODEs:
--- Trying classification methods ---
trying a quadrature
<- quadrature successful
<- 2nd order ODE linearizable_by_differentiation successful
-------------------
* Tackling next ODE.
*** Sublevel 3 ***
Methods for second order ODEs:
--- Trying classification methods ---
trying 2nd order Liouville
trying 2nd order WeierstrassP
trying 2nd order JacobiSN
differential order: 2; trying a linearization to 3rd order
trying 2nd order ODE linearizable_by_differentiation
<- 2nd order ODE linearizable_by_differentiation successful`
Maple dsolve solution
Solving time : 0.196 (sec)
Leaf size : 27
dsolve(diff(diff(y(x),x),x)^2+diff(y(x),x) = 0,y(x),singsol=all)
\begin{align*}
y &= c_{1} \\
y &= -\frac {1}{12} x^{3}+\frac {1}{2} c_{1} x^{2}-c_{1}^{2} x +c_{2} \\
\end{align*}
Mathematica DSolve solution
Solving time : 0.023 (sec)
Leaf size : 69
DSolve[{(D[y[x],{x,2}])^2+D[y[x],x]==0,{}},y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to -\frac {x^3}{12}-\frac {1}{4} i c_1 x^2+\frac {c_1{}^2 x}{4}+c_2 \\
y(x)\to -\frac {x^3}{12}+\frac {1}{4} i c_1 x^2+\frac {c_1{}^2 x}{4}+c_2 \\
\end{align*}