Solve the ODE $xy' = y-x\cos ^2\left(\dfrac{y}{x}\right)$.
I have tried as below. I'm using exact/non exact method.
We change the ODE into form
$$\left(y-x\cos ^2\left(\dfrac{y}{x}\right)\right) dx-xdy = 0.$$
Let $M(x,y)=y-x\cos ^2\left(\dfrac{y}{x}\right)$ and $N(x,y)=-x$.
\begin{align*}
\dfrac{\partial M(x,y)}{\partial y}&=
\dfrac{\partial }{\partial y}\left(y-x\cos ^2\left(\dfrac{y}{x}\right)\right) =
1+2\cos\left(\dfrac{y}{x}\right)\sin \left(\dfrac{y}{x}\right).\\
\dfrac{\partial N(x,y)}{\partial x}&=
\dfrac{\partial }{\partial x}(-x) = -1.
\end{align*}
Since $\dfrac{\partial M(x,y)}{\partial y}\neq \dfrac{\partial N(x,y)}{\partial x}$ then that ODE is non exact ODE.
Now I want to change it into exact ODE. We must find integrating factor to make the exact ODE. But now I don't know how to do it.
\begin{align*}
\dfrac{\partial M(x,y)}{\partial y}
–
\dfrac{\partial N(x,y)}{\partial x}
=
2+2\cos\left(\dfrac{y}{x}\right)\sin \left(\dfrac{y}{x}\right)
\end{align*}
Anyone can help me?
Best Answer
$$xy' = y-x\cos ^2\left(\dfrac{y}{x}\right)$$ The DE is homogeneous: $$y' = \dfrac yx-\cos ^2\left(\dfrac{y}{x}\right)$$ Substitute $y=tx \implies y'=t'x+t$ $$t'x=- \cos ^2 t$$ It's now separable. $$\int \dfrac {dt}{ \cos^2 t}=-\int \dfrac {dx}{x}$$