Answer:
Step-by-step explanation:
Given is a differential equation
[tex]y''-3y'+2y=3e^{2t}[/tex]
Take Laplace on both the sides
[tex]L(y")-3L(y')+2L(y) = L(e^{2t}[/tex]
[tex]s^2 Y(s)-sy(0) -y'(0) -3sY(s)+3y(0)+2Y(s)= L(3e^{2t} )\\Y(s)(s^2-3s+2)-1=\frac{3}{s-3}[/tex]
[tex]Y(s)(s^2-3s+2)=\frac{3}{s-3}+1\\Y(s) = \frac{s}{(s-3)(s-1)(s-2)}[/tex]
Resolve into partial fractions
[tex]Y(s) = \frac{-2}{(s-2)} +\frac{1}{2(s-1)} +\frac{3}{2(2-3)}[/tex]
Taking inverse
y(t) = [tex]-2e^{2t} +0.5e^t+1.5e^{3t}[/tex]
