Calculation of Integrating factor for non exact differential equation.
Differential equation $ydx+(2x-ye^y)dy=0$ has an integrating factor $\mu(x,y)=x^my^n$ for constants $m$ and $n$. Determine $\mu(x,y)$.
I tried to solve using the formula $M\frac {\partial\mu}{\partial y}-N\frac {\partial\mu}{\partial x}=(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\mu$. But still I didn't managed to find the $\mu(x,y)$.
The answer says $\mu(x,y)=y$.
But I want to know the steps to find this. I can solve the differential equation with $\mu(x,y)=y$. So please help me for finding the integrating factor $\mu(x,y)$.
$\endgroup$4 Answers
$\begingroup$Another method.
Since you know the integrating factor $\mu$ is of the form $x^{m}y^{n}$, multiply the differential equation by $x^{m}y^{n}$ to obtain $$x^{m}y^{n+1}dx+(2y^{n}x^{m+1}-x^{m}y^{n+1}e^{y})dy=0$$
For the differential equation to be exact we want $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
that is $$(n+1)y^{n}x^{m}=2(m+1)y^{n}x^{m}-mx^{m-1}y^{n+1}e^{y}$$
Since there are no $e^{y}$ terms on the LHS we choose $m=0$ which gives $$(n+1)y^{n}x^{0}=2y^{n}x^{0}$$ $$(n+1)y^{n}=2y^{n}$$
Thus $n=1$ and we have $\mu(x,y)=y$ as required.
See this also.
$\endgroup$ $\begingroup$Because you already know that $\mu = x^m y^n$, you can simply plug everything in to that formula. $M$ is equal to $y$, $N$ is equal to $2x-ye^y$, so the equation is $$y\left(nx^my^{n-1}\right)-\left(2x-ye^y\right)\left( mx^{m-1}y^n \right)=(2-1)x^my^n$$
Grouping into like terms makes it $$(n-1-2m)x^m y^n + me^y x^{m-1}y^{n+1} = 0$$
For it to be $0$ for all $x, y$, each term must be $0$. Therefore, $n-1-2m=0$ and $m = 0$. This then means that $y = 1$, and the answer is $$x^0 y^1 = y$$
$\endgroup$ $\begingroup$Letting $M=y, N=2x−ye^y$ in$$M\frac {\partial\mu}{\partial y}-N\frac {\partial\mu}{\partial x}=(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})\mu$$one has$$y\frac {\partial\mu}{\partial y}-(2x−ye^y)\frac {\partial\mu}{\partial x}=\mu.$$If $\mu$ is a function of $x$, one obtains$$ -(2x−ye^y)\frac {\partial\mu}{\partial x}=\mu $$which contains $y$; namely $\mu$ can't be a function of $y$. If $\mu$ is a function of $y$, one obtains$$ y\frac {\partial\mu}{\partial y}=\mu $$which can be solved easily to get $\mu=Cy$.
$\endgroup$ $\begingroup$$$ydx+(2x-ye^y)dy=0$$Another way:$$(ydx+2xdy)-ye^ydy=0$$Multiply by $y$:$$(y^2dx+xdy^2)-y^2e^ydy=0$$$$d(xy^2)-y^2e^ydy=0$$$$xy^2 =\int y^2e^ydy$$Integrate.
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