Internal
problem
ID
[18530] Book
:
Elementary
Differential
Equations.
By
Thornton
C.
Fry.
D
Van
Nostrand.
NY.
First
Edition
(1929) Section
:
Chapter
1.
section
5.
Problems
at
page
19 Problem
number
:
18 Date
solved
:
Saturday, February 22, 2025 at 09:21:08 PM CAS
classification
:
[_linear]
An ode of the form \(v' = \frac {M(u,v)}{N(u,v)}\) is called homogeneous if the functions \(M(u,v)\) and \(N(u,v)\) are both homogeneous functions and of the same order. Recall that a function \(f(u,v)\) is homogeneous of order \(n\) if
\[ f(t^n u, t^n v)= t^n f(u,v) \]
In this case, it can be seen that both \(M=-2 v +3 u\) and \(N=u\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {v}{u}\), or \(v=uu\). Hence
\[
-\frac {\ln \left (u \left (u \right )-1\right )}{3}=\ln \left (u \right )+c_1
\]
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
\[
-3 u +3=0
\]
for \(u \left (u \right )\) gives
\begin{align*} u \left (u \right )&=1 \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 {\ln \left (u \left (u \right )-1\right )}{3} &= \ln \left (u \right )+c_1 \\
u \left (u \right ) &= 1 \\
\end{align*}
Solving for \(u \left (u \right )\) gives
\begin{align*}
u \left (u \right ) &= 1 \\
u \left (u \right ) &= \frac {u^{3}+{\mathrm e}^{-3 c_1}}{u^{3}} \\
\end{align*}
Converting \(u \left (u \right ) = 1\) back to \(v\) gives
\begin{align*} v = u \end{align*}
Converting \(u \left (u \right ) = \frac {u^{3}+{\mathrm e}^{-3 c_1}}{u^{3}}\) back to \(v\) gives
\begin{align*} v = \frac {u^{3}+{\mathrm e}^{-3 c_1}}{u^{2}} \end{align*}
\[
-\frac {\ln \left (u \left (u \right )-1\right )}{3}=\ln \left (u \right )+c_1
\]
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
\[
-3 u +3=0
\]
for \(u \left (u \right )\) gives
\begin{align*} u \left (u \right )&=1 \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 {\ln \left (u \left (u \right )-1\right )}{3} &= \ln \left (u \right )+c_1 \\
u \left (u \right ) &= 1 \\
\end{align*}
Solving for \(u \left (u \right )\) gives
\begin{align*}
u \left (u \right ) &= 1 \\
u \left (u \right ) &= \frac {u^{3}+{\mathrm e}^{-3 c_1}}{u^{3}} \\
\end{align*}
Converting \(u \left (u \right ) = 1\) back to \(v\) gives
\begin{align*} v = u \end{align*}
Converting \(u \left (u \right ) = \frac {u^{3}+{\mathrm e}^{-3 c_1}}{u^{3}}\) back to \(v\) gives
\begin{align*} v = \frac {u^{3}+{\mathrm e}^{-3 c_1}}{u^{2}} \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 \(M=3 X -2 Y\) and \(N=X\) are both homogeneous and of the same order \(n=1\). 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
\[
3-3 u=0
\]
for \(u \left (X \right )\) gives
\begin{align*} u \left (X \right )&=1 \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.
We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that
If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is
Since \(\frac {\partial M}{\partial v} \neq \frac {\partial N}{\partial u}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let
\(M\) and \(N\) are multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) for now so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= u^{2}\left (\frac {2 v}{u}-3\right ) \\ &= -3 u^{2}+2 u v \end{align*}
Where \(f(u)\) is used for the constant of integration since \(\phi \) is a function of both \(u\) and \(v\). Taking derivative of equation (3) w.r.t \(u\) gives
Where \(c_1\) is constant of integration. Substituting result found above for \(f(u)\) into equation (3) gives \(\phi \)
\[
\phi = -u^{3}+v \,u^{2}+ c_1
\]
But since \(\phi \) itself is a constant function, then
let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as
\[
c_1 = -u^{3}+v \,u^{2}
\]
Solving for \(v\) gives
\begin{align*}
v &= \frac {u^{3}+c_1}{u^{2}} \\
\end{align*}
\[
-\frac {\ln \left (u \left (u \right )-1\right )}{3}=\ln \left (u \right )+c_1
\]
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
\[
-3 u +3=0
\]
for \(u \left (u \right )\) gives
\begin{align*} u \left (u \right )&=1 \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 {\ln \left (u \left (u \right )-1\right )}{3} &= \ln \left (u \right )+c_1 \\
u \left (u \right ) &= 1 \\
\end{align*}
Solving for \(u \left (u \right )\) gives
\begin{align*}
u \left (u \right ) &= 1 \\
u \left (u \right ) &= \frac {u^{3}+{\mathrm e}^{-3 c_1}}{u^{3}} \\
\end{align*}
Converting \(u \left (u \right ) = 1\) back to \(v\) gives
\begin{align*} \frac {v}{u} = 1 \end{align*}
Converting \(u \left (u \right ) = \frac {u^{3}+{\mathrm e}^{-3 c_1}}{u^{3}}\) back to \(v\) 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 &= u \\
\eta &= u \\
\end{align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( u,v\right ) \to \left ( R,S \right )\) 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 \(\left ( \xi \frac {\partial }{\partial u} + \eta \frac {\partial }{\partial v}\right ) S(u,v) = 1\). 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
\begin{align*} S &= \int { \frac {du}{T}}\\ &= \ln \left (u \right ) \end{align*}
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
We now need to express the RHS as function of \(R\) only. This is done by solving for \(u,v\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives
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 \(f(R)\).