Internal
problem
ID
[8129] Book
:
Second
order
enumerated
odes Section
:
section
2 Problem
number
:
40 Date
solved
:
Monday, October 21, 2024 at 04:54:06 PM CAS
classification
:
[_rational]
Solve
\begin{align*} 2 x y^{2}-y+\left (y^{2}+x +y\right ) y^{\prime }&=0 \end{align*}
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. Since the ODE is not exact, we will try to find an
integrating factor to make it exact. Let
\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=\frac {1}{y^{2}+x +y}\left ( \left ( 4 y x -1\right ) - \left (1 \right ) \right ) \\ &=\frac {4 y x -2}{y^{2}+x +y} \end{align*}
Since \(A\) depends on \(y\), it can not be used to obtain an integrating factor. We will now try a
second method to find an integrating factor. Let
\(M\) and \(N\) are now multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\)
and \(\overline {N}\) so not to confuse them with the original \(M\) and \(N\).
\begin{align*} \overline {M} &=\mu M \\ &= \frac {1}{y^{2}}\left (2 x \,y^{2}-y\right ) \\ &= \frac {2 y x -1}{y} \end{align*}
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