I’m not able to find an integrating factor for the next ordinary differential equation:
$$ x + (1-y^{1/2}) \tan{(x-y)} = \left[ x-y^{1/2} \tan{(x-y)} - \frac{1}{2} y^{-1/2} \right] y' $$
I know that it must be of the type $m(x)$, $m(y)$, $m(xy)$, or $m(x±y)$. Could anyone help me a little with this?
$$ \left( x + (1-y^{1/2}) \tan{(x-y)}\right)dx - \left( x-y^{1/2} \tan{(x-y)} - \frac{1}{2} y^{-1/2} \right) dy =0$$ In order to simplify the search of integrating factor, change of variable : $\quad t=x-y \quad\to\quad x=y+t \quad\to\quad dx=dy+dt$
$$ \left( y+t + (1-y^{1/2}) \tan(t)\right)(dy+dt) - \left( y+t-y^{1/2} \tan(t) - \frac{1}{2} y^{-1/2} \right) dy =0$$ After simplification : $$ \left(\tan(t) + \frac{1}{2} y^{-1/2} \right)dy +\bigg(y+t-(1-\sqrt{y})\tan(t) \bigg)dt =0$$
$$ \left(\sin(t) + \frac{1}{2} y^{-1/2}\cos(t) \right)dy +\bigg((y+t)\cos(t)-(1-\sqrt{y})\sin(t) \bigg)dt =0$$
$$\frac{\partial}{\partial x} \left(\sin(t) + \frac{1}{2} y^{-1/2}\cos(t) \right) =\frac{\partial}{\partial y}\bigg((y+t)\cos(t)-(1-\sqrt{y})\sin(t) \bigg)=$$ $$=\cos(t)-\frac{1}{2} y^{-1/2}\sin(t) $$
Hence $ \left(\sin(t) + \frac{1}{2} y^{-1/2}\cos(t) \right)dy +\big((y+t)\cos(t)-(1-\sqrt{y})\sin(t) \big)dt $ is a total differential.
$\begin{cases} \int \left(\sin(t) + \frac{1}{2} y^{-1/2}\cos(t) \right)dy =y\sin(t)+\sqrt{y}\cos(t)+f(t) \\ \int \big((y+t)\cos(t)-(1-\sqrt{y})\sin(t) \big)dt= y\sin(t)+\sqrt{y}\cos(t)+t\sin(t)+g(y) \end{cases}$
which implies $\quad g(y)=c=$ constant$\quad$ and $\quad f(t)=t\sin(t)+c$.
The solution is expressed on the form of an implicit equation : $$y\sin(t)+\sqrt{y}\cos(t)+t\sin(t)=C$$ $$y\sin(x-y)+\sqrt{y}\cos(x-y)+(x-y)\sin(x-y)=C$$ $$\sqrt{y}\cos(x-y)+x\sin(x-y)=C$$