How do you solve the separable differential equation #dy/dx=(cosx)e^(y-sinx)#?

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Answer 1 · 2017-03-10T18:30:22

The answer is #y=-ln(e^-sinx-C)#

We need

#e^(a-b)=e^(a)/e^b#

Let's rewrite and simplify the equation

#dy/dx=cosx e^(y-sinx)#

#dy/dx=cosx e^y/e^sinx#

#dy/e^y=cosx/e^sinxdx#

Integrating both sides

#intdy/e^y=intcosx/e^sinxdx#

#intdy/e^y=-e^-y#

For the #RHS#, we perform a substitution

Let #u=sinx#, #=>#, #du=cosx dx#

#intcosx/e^sinxdx=int(du)/e^(u)#

#=-e^(-u)#

#=-e^(-sinx)#

Therefore,

#-e^-y=-e^(-sinx)+C#

#e^-y=e^(-sinx)-C#

#-y=ln(e^-sinx-C)#

#y=-ln(e^-sinx-C)#