Question

Use Separation of variables to solve the initial value problem: dy/dx=x-3+xy-3y; y(0)=3

Answer 1

dy/dx = x-3 +y(x-3)...
dy/dx = (x-3) +y(x-3)...
dy/dx = (x-3)(1+y)
dy = (x-3)(1+y)dx
(1/(1+y))dy = (x-3)dx
integrate each side...
ln(1+y) = ((x^2)/2) -3x + C...
solve for y...
e^(1+y) = (e^((x^2)/2) - 3x) + K...
1+y = e^((x^2/2))-3x) + K...

Answer 2

y(x) = 4e^((1/2)(x-6)x) -1 (using given initial condition).

Answer 3

*note that -1 is not part of the exponent.