seperation by diff equations?

Question

Find the particular solution of the differential equation
{dy}/{dx} = (x-4)e^{-2y}
satisfying the initial condition y(4)=ln(4).
Answer: y= .

Answer 1

dy/e^-2y = (x-4)dx
∫e^2ydy = x^2 - 4x + C
let u = 2y
∫e^udu =2x^2 - 8x + C
e^2y = 2x^2 - 8x + C
2y = ln(2x^2 - 8x + C)
y = (1/2)ln(2x^2 - 8x + C)
y(4) = ln(4)
2*16 - 32 + C = 16
C = 16
y = (1/2)ln(2x^2 - 8x + 16)