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

Answer:

Step-by-step explanation:

Given is a differential equation

$y''-3y'+2y=3e^{2t}$

Take Laplace on both the sides

$L(y")-3L(y')+2L(y) = L(e^{2t}$

$s^2 Y(s)-sy(0) -y'(0) -3sY(s)+3y(0)+2Y(s)= L(3e^{2t} )\\Y(s)(s^2-3s+2)-1=\frac{3}{s-3}$

$Y(s)(s^2-3s+2)=\frac{3}{s-3}+1\\Y(s) = \frac{s}{(s-3)(s-1)(s-2)}$

Resolve into partial fractions

$Y(s) = \frac{-2}{(s-2)} +\frac{1}{2(s-1)} +\frac{3}{2(2-3)}$

Taking inverse

y(t) = $-2e^{2t} +0.5e^t+1.5e^{3t}$