Question

Find the solution to the given IVP using the Laplace transform technique: y''-3y'+2y=3e^(2t), y(0)=0; y'(0)=1

Answer 1

Explanation:

Given is a differential equation

`y''-3y'+2y=3e^(2t)`

Take Laplace on both the sides

`L(y`

`s^2 Y(s)-sy(0) -y'(0) -3sY(s)+3y(0)+2Y(s)= L(3e^(2t) )\\Y(s)(s^2-3s+2)-1=(3)/(s-3)`

`Y(s)(s^2-3s+2)=(3)/(s-3)+1\\Y(s) = (s)/((s-3)(s-1)(s-2))`

Resolve into partial fractions

`Y(s) = (-2)/((s-2)) +(1)/(2(s-1)) +(3)/(2(2-3))`

Taking inverse

y(t) =
`-2e^(2t) +0.5e^t+1.5e^(3t)`