Internal
problem
ID
[7714] Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
23 Date
solved
:
Monday, October 21, 2024 at 03:58:25 PM CAS
classification
:
[[_homogeneous, `class A`], _rational, [_Abel, `2nd type`, `class B`]]
Solve
\begin{align*} x^{2} y^{\prime }+y^{2}&=x y y^{\prime } \end{align*}
1.22.1 Solved as first order homogeneous class A ode
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 \(M=-y^{2}\) and \(N=x \left (x -y \right )\) 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 \(\frac {u}{u -1}=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=0 \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.
Which is now solved The ode \(u^{\prime }\left (x \right ) = \frac {u \left (x \right )}{\left (u \left (x \right )-1\right ) x}\) is separable as it can be written as
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 \(\frac {u}{u -1}=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=0 \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.
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 \(M=-Y^{2}\) and \(N=X \left (X -Y \right )\) 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
Which is now solved as separable in \(u \left (X \right )\).
The ode \(\frac {d}{d X}u \left (X \right ) = \frac {u \left (X \right )}{X \left (u \left (X \right )-1\right )}\) is separable as it can be written as
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 \(\frac {u}{u -1}=0\) for \(u \left (X \right )\) gives
\begin{align*} u \left (X \right )&=0 \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 \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. By inspection \(\frac {1}{x^{2} y}\) is an integrating factor. Therefore by
multiplying \(M=y^{2}\) and \(N=x^{2}-y x\) by this integrating factor the ode becomes exact. The new \(M,N\) are
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 \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the
function \(\phi \left (x,y\right )\)
Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function
of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives
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
\[
c_1 = -\frac {y}{x}+\ln \left (y \right )
\]
Solving for \(y\) from the
above solution(s) gives (after possible removing of solutions that do not verify)
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 \(\frac {u}{u -1}=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=0 \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.
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 \(S(R)\). Since
\(\xi =0\) then in this special case
\begin{align*} R = x \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {x y}{x -y}}} dy \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} &= 0 \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 \(f(R)\).