Answer to Question #134077 in Differential Equations for Ajit Singh

Question

Solve the IVP:

y" + 3y' + 2y = 6((delta)base 6)(t)

with conditions: y(0) = 0, y'(0) =-1.

Answer 1

Given differential equation is

$y''+3y'+2y = \delta_6(t)$ with conditions is $y(0)=0, y'(0)=-1$ .

Taking Laplace transform on both sides, we get

$L(y'')+ 3L(y')+2L(y )= L(\delta_6(t)) \\ \implies s^2 Y + 1 + 3 sY + 2 Y = e^{-6s}$ where $Y = L(y)$ .

$\implies (s^2+3s+2) Y = e^{-6s} -1 \\ \implies Y =\frac{ e^{-6s} -1}{s^2+3s+2} \\ \implies y =L^{-1}(\frac{ e^{-6s} -1}{s^2+3s+2})$

So, $y =\left(\left(\left(- e^{12 - t} + e^{6}\right) \delta_6(t )- 1\right) e^{t} + 1\right) e^{- 2 t}$