Question

Question #a071a

Answer 1

`mu=-sinx*e^cosx`

Explanation:

First manipulate the equation back into standard form, `dy/dx +p(x)y=q(x)`

Your equation is
`dy/dx+ysinx=e^(x)`
`:.p(x) = sinx and q(x)=e^x`

`mu = e^(int(p(t))dt)`
`:. mu= e^(int(sinx)dx)`
`=e^cosx`
`=-sinx*e^cosx` (multiply by the derivative of the power)