Solving differential equation by exact method

Question

I am trying to solve this equation by method of reducing to exact equation ; i.e. $M(x,y)dx + N(x,y)dy = 0$

The equation is $dy/dx = \cos^2y / \sin^2x$

$∂M/∂y = 2\cos y \sin y$

$∂N/∂x = 2\cos x \sin x$

Clearly it is not exact. I tried making it exact using $μ(x)$, $μ(y)$ and $μ(xy)$ but to no avail.

Any insight would be helpful.

Answer 1

$$dy/dx = \cos^2(y) / \sin^2(x)$$ $$\frac {dy} {\cos^2(y)}-\frac {dx} {\sin^2(x)} =0$$ Rewrite it as: $$ d(\tan (y)) +d(\text {ctg} (x)) =0$$ It's exact. $$ Nd(\tan (y)) +Md(\text {ctg} (x)) =0$$ $$M(\tan (y),\text {ctg} (x))=N(\tan (y),\text {ctg} (x))=1$$

Answer 2

You wrote : $$M(x,y)dx + N(x,y)dy = 0$$

$$dy/dx = \cos^2y / \sin^2x$$

$$∂M/∂y = 2\cos y \sin y$$

$$∂N/∂x = 2\cos x \sin x$$

This is false because you made a mistake in the calculus of $N$ and $M$

$$dy/dx = \cos^2y / \sin^2x\quad\implies\quad -\frac{dx}{\sin^2(x)}+\frac{dy}{\cos^2(y)}=0 $$ Now this is the correct form $\quad M(x,y)dx + N(x,y)dy = 0\quad$ thus : $$M(x,y)=-\frac{1}{\sin^2(x)}$$ $$N(x,y)=\frac{1}{\cos^2(y)}$$ As a consequence $$∂M/∂y =∂N/∂x = 0$$

This is the simplest case of $∂M/∂y =∂N/∂x$ .

Thus $\quad-\frac{dx}{\sin^2(x)}+\frac{dy}{\cos^2(y)}=0\quad$ is exact.

The integrating factor is $\mu=1$ as is is wellknown for all exact DE.

So, one can directly integrate it for : $$\cot(x)+\tan(y)=c$$ $$y(x)=\tan^{-1}\left(c-\cot(x) \right)$$

Answer 3

The equation $$dy/dx = \cos^2y / \sin^2x$$ is exact if you write it as $$\csc ^2 x dx-\sec ^2 y dy =0$$