Internal
problem
ID
[6397] Book
:
Ordinary
differential
equations
and
calculus
of
variations.
Makarets
and
Reshetnyak.
Wold
Scientific.
Singapore.
1995 Section
:
Chapter
1.
First
order
differential
equations.
Section
1.1
Separable
equations
problems.
page
7 Problem
number
:
34 Date
solved
:
Thursday, October 17, 2024 at 10:07:16 AM CAS
classification
:
[_separable]
We now need to find the singular solutions, these are found by finding for what values \(g(y)\) is
zero, since we had to divide by this above. Solving \(g(y)=0\) or \(\frac {y^{2}}{y -1}=0\) for \(y\) gives
\begin{align*} y&=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
\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (x \,y^{2}+y^{2}\right )\\ &= 2 x y +2 y \end{align*}
And
\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (-x^{2} y +x^{2}\right )\\ &= -2 x y +2 x \end{align*}
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}{\left (y -1\right ) x^{2}}\left ( \left ( 2 x y +2 y\right ) - \left (-2 x y +2 x \right ) \right ) \\ &=\frac {\left (-4 y +2\right ) x -2 y}{\left (y -1\right ) x^{2}} \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
\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} \right ) \\ &=\frac {1}{y^{2} \left (x +1\right )}\left ( \left ( -2 x y +2 x\right ) - \left (2 x y +2 y \right ) \right ) \\ &=\frac {\left (-4 y +2\right ) x -2 y}{y^{2} \left (x +1\right )} \end{align*}
Since \(B\) depends on \(x\), it can not be used to obtain an integrating factor.We will now try a third
method to find an integrating factor. Let
\[ R = \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \]
\(R\) is now checked to see if it is a function of only \(t=xy\).
Therefore
\begin{align*} R &= \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \\ &= \frac {\left (-2 x y +2 x\right )-\left (2 x y +2 y\right )} {x\left (x \,y^{2}+y^{2}\right ) - y\left (-x^{2} y +x^{2}\right )} \\ &= -\frac {2}{x y} \end{align*}
Replacing all powers of terms \(xy\) by \(t\) gives
\[ R = -\frac {2}{t} \]
Since \(R\) depends on \(t\) only, then it can be used to find
an integrating factor. Let the integrating factor be \(\mu \) then
Multiplying \(M\) and \(N\) by this integrating factor gives new \(M\)
and new \(N\) which are called \( \overline {M}\) and \( \overline {N}\) so not to confuse them with the original \(M\) and \(N\)
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
Solving for \(y\) from the
above solution(s) gives (after possible removing of solutions that do not verify)
\begin{align*} y = {\mathrm e}^{\frac {\operatorname {LambertW}\left (-{\mathrm e}^{-\frac {\ln \left (x \right ) x -c_1 x -1}{x}}\right ) x +\ln \left (x \right ) x -c_1 x -1}{x}} \end{align*}
`Methodsfor first order ODEs:---Trying classification methods ---tryinga quadraturetrying1st order lineartryingBernoullitryingseparable<-separable successful`