Question

dy/dt + 4y = 6e ^2t
With y (t=0) = 3, Use laplace transform

Answer 1

Let us take Laplace transform of both sides of differential equation, using property of Laplace transform $L[y'] = s L[y] - y(0)$ and table Laplace transform $L[e^{a t}] = \frac{1}{s-a}$ :

$s Y(s) - 3 + 4Y(s) = \frac{6}{s-2}$. Simplifying and solving for $Y(s)$, obtain:

$Y(s)[s+4] = \frac{6}{s-2} + 3 = \frac{3 s}{s-2} \Rightarrow Y(s) = \frac{3 s}{(s+4)(s-2)}$.

In order to take inverse Laplace transform, decompose the last expression into partial fractions:$\frac{3 s}{(s+4)(s-2)} = \frac{2}{s+4} + \frac{1}{s-2}$. Hence, using table Laplace transform $L[e^{a t}] = \frac{1}{s-a}$, obtain $y(t) = L^{-1}[Y(s)] = L^{-1}[\frac{2}{s+4} + \frac{1}{s-2}] = 2 e^{-4 t} + e^{2 t}$.

It is straightforward to check that this is indeed the solution to given differential equation.