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[8742] Book
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Own
collection
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miscellaneous
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section
1.0 Problem
number
:
31 Date
solved
:
Friday, February 21, 2025 at 08:26:57 PM CAS
classification
:
[[_homogeneous, `class A`], _rational, _Riccati]
An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if
\[ f(t^n x, t^n y)= t^n f(x,y) \]
In this case, it can be seen that both \(M=5 x^{2}-x y +y^{2}\) and \(N=x^{2}\) are both homogeneous and of the same order \(n=2\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence
We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or
\[
u^{2}-2 u +5=0
\]
for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=1-2 i\\ u \left (x \right )&=1+2 i \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.
Therefore the solutions found are
\begin{align*}
\frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} &= \ln \left (x \right )+c_1 \\
u \left (x \right ) &= 1-2 i \\
u \left (x \right ) &= 1+2 i \\
\end{align*}
Solving for \(u \left (x \right )\) gives
\begin{align*}
u \left (x \right ) &= 1-2 i \\
u \left (x \right ) &= 1+2 i \\
u \left (x \right ) &= 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) \\
\end{align*}
Converting \(u \left (x \right ) = 1-2 i\) back to \(y\) gives
\begin{align*} y = \left (1-2 i\right ) x \end{align*}
Converting \(u \left (x \right ) = 1+2 i\) back to \(y\) gives
\begin{align*} y = \left (1+2 i\right ) x \end{align*}
We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or
\[
u^{2}-2 u +5=0
\]
for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=1-2 i\\ u \left (x \right )&=1+2 i \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.
Therefore the solutions found are
\begin{align*}
\frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} &= \ln \left (x \right )+c_1 \\
u \left (x \right ) &= 1-2 i \\
u \left (x \right ) &= 1+2 i \\
\end{align*}
Solving for \(u \left (x \right )\) gives
\begin{align*}
u \left (x \right ) &= 1-2 i \\
u \left (x \right ) &= 1+2 i \\
u \left (x \right ) &= 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) \\
\end{align*}
Converting \(u \left (x \right ) = 1-2 i\) back to \(y\) gives
\begin{align*} y = \left (1-2 i\right ) x \end{align*}
Converting \(u \left (x \right ) = 1+2 i\) back to \(y\) gives
\begin{align*} y = \left (1+2 i\right ) x \end{align*}
An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if
\[ f(t^n X, t^n Y)= t^n f(X,Y) \]
In this case, it can be seen that both \(M=5 X^{2}-X Y +Y^{2}\) and \(N=X^{2}\) are both homogeneous and of the same order \(n=2\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence
We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or
\[
u^{2}-2 u +5=0
\]
for \(u \left (X \right )\) gives
\begin{align*} u \left (X \right )&=1-2 i\\ u \left (X \right )&=1+2 i \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.
Therefore the solutions found are
\begin{align*}
\frac {\arctan \left (\frac {u \left (X \right )}{2}-\frac {1}{2}\right )}{2} &= \ln \left (X \right )+c_1 \\
u \left (X \right ) &= 1-2 i \\
u \left (X \right ) &= 1+2 i \\
\end{align*}
Solving for \(u \left (X \right )\) gives
\begin{align*}
u \left (X \right ) &= 1-2 i \\
u \left (X \right ) &= 1+2 i \\
u \left (X \right ) &= 1+2 \tan \left (2 \ln \left (X \right )+2 c_1 \right ) \\
\end{align*}
Converting \(u \left (X \right ) = 1-2 i\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = \left (1-2 i\right ) X \end{align*}
Converting \(u \left (X \right ) = 1+2 i\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = \left (1+2 i\right ) X \end{align*}
We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or
\[
u^{2}-2 u +5=0
\]
for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=1-2 i\\ u \left (x \right )&=1+2 i \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.
Therefore the solutions found are
\begin{align*}
\frac {\arctan \left (\frac {u \left (x \right )}{2}-\frac {1}{2}\right )}{2} &= \ln \left (x \right )+c_1 \\
u \left (x \right ) &= 1-2 i \\
u \left (x \right ) &= 1+2 i \\
\end{align*}
Solving for \(u \left (x \right )\) gives
\begin{align*}
u \left (x \right ) &= 1-2 i \\
u \left (x \right ) &= 1+2 i \\
u \left (x \right ) &= 1+2 \tan \left (2 \ln \left (x \right )+2 c_1 \right ) \\
\end{align*}
Converting \(u \left (x \right ) = 1-2 i\) back to \(y\) gives
\begin{align*} \frac {y}{x} = 1-2 i \end{align*}
Converting \(u \left (x \right ) = 1+2 i\) back to \(y\) gives
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives
\begin{align*}
\xi &= x \\
\eta &= y \\
\end{align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Therefore
Where the constant of integration is set to zero as we just need one solution. Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= \frac {x^{2}}{5 x^{2}-2 x y +y^{2}}\tag {2A} \end{align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives
\begin{align*} \frac {dS}{dR} &= \frac {1}{R^{2}-2 R +5} \end{align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).
This is Euler second order ODE. Let the solution be \(u = x^r\), then \(u'=r x^{r-1}\) and \(u''=r(r-1) x^{r-2}\). Substituting these back into the given ODE gives
\[ x^{2}(r(r-1))x^{r-2}+3 x r x^{r-1}+5 x^{r} = 0 \]
Using Euler relation, the expression \(c_1 e^{i A}+ c_2 e^{-i A}\) is transformed to \( c_1 \cos A+ c_1 \sin A\) where the constants are free to change. Applying this to the above result gives
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparabletryinginverse lineartryinghomogeneous types:tryinghomogeneous D<-homogeneous successful`