Question

What is a solution to the differential equation `x(y+4)+y'=0` with y=-5 when x=0?

Answer 1

`y = - e^(-x^2/2 ) - 4`

Explanation:

`x(y+4)+y'=0`

this is separable

`y' = - x(y+4)`

`1/(y+4) y' = - x`

`int \ 1/(y+4) y' \ dx = - int \ x \ dx`

`int \ 1/(y+4) y' \ dx = - int \ x \ dx`

`ln(y+4) = -x^2/2 + C`

`y+4 = e^(-x^2/2 + C)`

`y = C e^(-x^2/2 ) - 4`

Applying the IV `y(0) = -5`

`implies -5 = C e^(0) - 4`

`implies C = -1`

`y = - e^(-x^2/2 ) - 4`