Dy/dx = (cos x) e^(y+sinx) and y = 0 when x = 0. find the original equation

Find the general solution by separating the variables then integrating: 
dy / dx = cosx℮^(y + sinx) 
dy / dx = cosx℮ʸ℮^(sinx) 
℮^(-y) dy = cosx℮^(sinx) dx 
∫ ℮^(-y) dy = ∫ cosx℮^(sinx) dx 
-℮^(-y) = ℮^(sinx) + C 
℮^(-y) = C - ℮^(sinx) 
-y = ln[C - ℮^(sinx)] 
y = -ln[C - ℮^(sinx)] 

Find the particular solution by solving for the constant: 
When x = 0, y = 0 
-ln(C - 1) = 0 
ln(C - 1) = 0 
C - 1 = 1 
C = 2 
y = -ln[2 - ℮^(sinx)]

I hope my answer has come to your help. Thank you for posting your question here in . We hope to answer more of your questions and inquiries soon. Have a nice day ahead!