Differential Equations?

Question

I'm confused by this... maybe someone else can get it.

Use seperation of variables to solve the initial value problem


dy/dx = cos(x) * e^(y+sin x)

x = 0 , y = 0

So far I have 1/e^y * dy = cos x * e^sin(x) dx

Could I just say 1/e^y is a constant, so C * dy = cos x * e^ sin(x) dx, then integrate, so C = ∫ (cos x * e ^ sin (x)) dx ?

Solve for in terms of y.

Answer 1

dy/dx=cosx*e^y*e^sinx
dy/e^y= cosx*e^sinx*dx
e^-ydy= cosx*e^sinx*dx
integrate
e^-y/(-1)=int. cosx*e^sinx*dx
put sinx=t
cosxdx=dt
-e^-y=int.e^tdt
=e^t+c
=e^sinx+c
put x=0,y=0
-1=1+c
c=-2
-e^-y=e^sinx-2

Answer 2

You have done all the hard yards:

1/e^y * dy = cos x * e^sin(x) dx
So ∫e^(-y)dy = ∫e^sin(x) .cos(x) dx
ie -e^(-y) = e^sin(x) + c

So e^(-y) = K - e^sin(x)
Taking lns
-y = ln (K - e^sin(x))
ie y = -ln (K - e^sin(x))
= ln (1/(K - e^sin(x))
When x = 0 y = 0
So 0 = ln (1/(K - 1))
1/(K - 1) = 1
K = 2
Thus y = ln (1/(2 - e^sin(x)))
{= -ln (2 - e^sin(x))}

Check dy/dx = - -cos(x) e^sin(x)/(2 - e^sin(x))
= cos(x) e^sin(x)/e^(-y)
= cos(x)*e^(y + sin(x)) ... QED

Answer 3

dy/dx = cos(x) * e^(y+sin x)
dy/dx = cos(x) * e^y*e^(sin x)
(e^-y)dy = (cos(x) * e^(sin x))dx
let u = sin(x), then du = cos(x)dx
∫(e^-y)dy = ∫(e^u)du
-e^-y = e^u = e^sin(x) + C
-1 = 1 + C
C = -2
e^-y = 2 - e^sin(x)
e^y = 1/(2 - e^sin(x))
y = ln[1/(2 - e^sin(x))], -3.907 + 2nπ < x < 0.766 + 2nπ