## anonymous 5 years ago Determine the solution to the initial value differential equation: 3y'-5y=e^(2x), y(0)=13

1. anonymous

Use undetermined coefficients or variation of paramaters

2. anonymous

Get the equation in the form y' -5/3*y=e^(2x)/3, then get the multiply both sides by the integrating factor e^(-5x/3) to get $(e ^{-5x/3}*y)'=1/3*e^{2x}*e^{-5x/3} \rightarrow (e ^{-5x/3}*y)' = 1/3*e^{x/3}$ All you have to do next is take the antiderivative of both sides and then divide both sides by $e^{-5x/3}$

3. anonymous

oh oops, nevermind what i said thought it was 3y'' lol jprahman is right