dy/dx=y² sin x , if y(0)=1.?

Question

Find the solution to the following seperable differential equation if y(0)=1

Answer 1

dy/dx=y² sin x
dy/y^2=sinxdx

integrate
-1/y = -cosx +K
1/y = cosx -K
use initial conditions
1=1-K
K=0
y= secx

remember, int{sin}= -cos

i hope that this helps

Answer 2

∫dy/y² = ∫sinx dx
-1/y = cosx + c
y = -1/cosx + c
y(0) = -1/cos 0 + c = -1 + c = 1, so c = 2.

So y(x) = 2 - 1/cosx

Disclaimer: it's been YEARS since I did this, literally. I could be totally off.

Answer 3

dy/y^2 = sin(x) dx

integrate both sides

-1/y = cos(x) + c
y = -1/cos(x) +c

when x = 0, y(0) = 1

1 = -1/cos(0) +c
1+1/1 = c
c = 2

so the final answer is

y = 2 - 1/cos(x)

Answer 4

dy/dx = y² sin(x)
∫dy/y² = ∫sin(x)dx
-1/y = -cos(x) - c
y = 1/(cos(x) + c )
y(0) = 1 = 1/(1+c)
c+1 = 1
c = 1
y(x) = 1/cos(x)

Answer 5

No solution,no answer