Solve the following IVP with explicit solution

Question

Given:

$4 dx + 2 {cos(y)\over sin(y)} dy = 0, \qquad y(0) = {\pi\over 2}$

I've already test the exactness which is $0$ for the result of both derivatives.

Then I found the potential function is $u(x,y) = 4x + 2 ln (sin(y))$

But I don't know how to find the explicit solution for $y(x)$

Because when I find $u(0,{\pi\over 2})$, the result is $0$

or do I have to find $y$ from $4x + 2 ln (sin(y)) = 0$ ? I thought it's an implicit solution.

Many thanks for your help.

Answer 1

$$4 dx + 2 {\cos(y)\over \sin(y)} dy = 0$$

Note that the ODE is sperable. We can simpliy integrate it:

$$4x+2\ln|\sin(y)|=c.$$

Apply IV (initial value):

$$4\cdot 0+2\ln|\sin(\pi/2)|=c \implies c=0.$$

$$2\ln|\sin(y)|=-4x \implies \ln|\sin(y)|=-2x \implies |\sin(y)|=\exp(-2x).$$

As our IV was $y=\pi/2$, we can assume $y \in (0,\pi)$. This implies that the absolute value is positive. Hence,

$$\sin(y)=\exp(-2x) \implies y = \arcsin\left[\exp(-2x)\right]$$

Answer 2

$$\frac{dx}{dy}=-\frac{1}{2}\frac{\cos y}{\sin y}$$ $$x(\pi/2)=0$$ $$x=-\frac{1}{2}\int \frac{\cos y}{\sin y}dy$$ $$=-\frac{1}{2}\ln(\sin y)+C$$ $$0=-\frac{1}{2}\ln(\sin \frac{\pi}{2})+C=C$$ $$x=-\frac{1}{2}\ln(\sin y)$$ $$y=\sin^{-1}(e^{-2x})$$