First of all, the question is not about the general solution. My question on my book asks me to find the integrating factor.
Suppose I have simple IDE
$$(\sin y)\mathbb dx +(\cos x)\mathbb dy=0$$
I don't know how to find the integrating factor $(\mu)$ manually.
What i've done so far is :
Suppose it has form
$$M(x,y) \mathbb dx+N(x,y) \mathbb dy=0$$
Case 1 : $\mu$ is a function of $x$ only
$$\begin{aligned} \mu&=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\\ &=\frac{\cos y +\sin x}{\cos x} \end{aligned}$$
It doesn't work, cz still remaining the $y$
Case 2 : $\mu$ is a function of $y$ only
$$\begin{aligned} \mu&=\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\\ &=\frac{-\sin x -\cos y}{\sin y} \end{aligned}$$
Still remaining the $x$
Case 3 : $\mu$ is a function of $xy$
$$\begin{aligned} \mu&=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{My - Nx}\\ &=\frac{\cos y +\sin x}{y \sin x-x \sin y} \end{aligned}$$
I have no idea for this one...
But, when i use trial and error, i found
$$\mu=\csc y\cdot \sec x$$
But it's just a guess.
How to find the integrating factor systematically? I mean using such $\mu=e^{\int f(x)\,\mathbb dx}$?
Help me, and thanks.
$$\sin y \,dx + \cos x dy =0 ~~~(1)\implies \frac{dx}{\cos x}+ \frac{dy}{\sin y}=0~~~(2).$$ Integrating, we get $$ \ln|\tan(x)+\sec(x)|- \ln|\cot(y)+\csc(y)|=C.$$
Note it is as though $IF = \dfrac{1}{\cos x \sin y}= \sec x \csc y,$ which upon multiplying the ODE (1) makes it EXACT as (2) where $$\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}.$$