Find function which satisfies the initial value problem: $\displaystyle 6 \sec x \,\frac{dy}{dx} = e^{y + \sin x} $

Question

Find the function which satisfies the initial value problem:

$\displaystyle 6 \sec x \,\frac{dy}{dx} = e^{y + \sin x} $

$\displaystyle \;\ y(0) = -9 $

So as far as I understand it, I should move everything with y to the left, and with x to the right.

But what do I do with $e^{y + \sin(x)}$? it has both y and x

if I leave it on the right side, do I just do $u$ substitution $= y + \sin(x)$?

Answer 1

Outline :

You can separate RHS by writing $e^ye^{\sin x}$.

Now, the equation is in separable form,

$6e^{-y}dy = \cos x \times e^{sin x} dx $

LHS is easily integrable, for RHS substitute $t = \sin x$, so you will get

$-6e^{-y} = e^{\sin x} + c$

You can get c from the initial condition

Answer 2

Hint:

$$e^{y+\sin x}=e^y\cdot e^{\sin x}\implies 6e^{-y}\frac{dy}{dx}=e^{\sin x}\cos x$$

Then, by integrating both sides, you get $$-6e^{-y}=e^{\sin x}+C$$