1. The problem statement, all variables and given/known data Find the general solution of the first-order differential equation, y' + 3y = 2xe^(-3x) 2. Relevant equations y' + P(x)y = Q(x) Integrating factor = e^(∫P(x) dx) 3. The attempt at a solution Since it's already in the form y' + P(x)y = Q(x), the integrating factor is I(x) = e^(∫3 dx) = e^(3x) Now multiplying both sides by the integrating factor, e^(3x)*y' + 3ye^(3x) = 2xe^(-3x)e^(3x) d/dx (e^(3x)y) = 2x Finally, integrating both sides and solving for y, e^(3x)y = x^2 + C General solution: y = x^(2)e^(-3x) + Ce^(-3x) Not to sure if this is right, this is the first time I've studied diff eqns.. Thanks for any help.