Question

Consider the initial value problem

y' + 6y = {0 if 0 < or equal to t < or equal to 2
12 if 2 < or equal to t < or equal to 6
0 if 6 < or equal to t < or equal to infinity}
a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t) by Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below).
b. Solve your equation for Y(s).
c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t).

Answer 1

y' + 6y = f(t)

where

`f(t)=\begin{cases}0&\text{if }0\le t\le2\\12&\text{if }2<t\le6\\0&\text{if }6<t<\infty\end{cases}`

You can write f(t) in terms of the unit step (i.e. Heaviside theta) function u(t), which is defined as

`u(t)=\begin{cases}0&\text{if }t<0\\1&\text{if }t\ge0\end{cases}`

Then the DE is written as

y' + 6y = 12 u (t - 2) - 12 u (t - 6)

(a) Take the Laplace transform of both sides:

LT[y' + 6y] = LT[12 u (t - 2) - 12 u (t - 6)]

s Y - y (0) + 6Y = 12 (exp(-2s) - exp(-6s))/s

(b) Solve for Y :

(s + 6) Y = 12 (exp(-2s) - exp(-6s))/s + y (0)

Y = 12 (exp(-2s) - exp(-6s))/(s (s + 6)) + y (0)/(s + 6)

(c) Take the inverse transform:

LT⁻¹ [Y] = LT⁻¹[12 (exp(-2s) - exp(-6s))/(s (s + 6)) + y (0)/(s + 6)]

y = 12 LT⁻¹ [(exp(-2s) - exp(-6s))/(s (s + 6))] + y (0) LT⁻¹ [1/(s + 6)]

y = 12 u (t - 2) LT⁻¹ [1/(s (s + 6))] - 12 u (t - 6) LT⁻¹ [1/(s (s + 6))] + y (0) exp(-6t )

For the remaining inverse transform, break up into partial fractions:

1/(s (s + 6)) = a/s + b/(s + 6)

1 = a (s + 6) + bs

1 = (a + b) s + 6a

==> 6a = 1, a + b = 0 ==> a = 1/6, b = -1/6

y = 2 u (t - 2) LT⁻¹ [1/s - 1/(s + 6)] - 2 u (t - 6) LT⁻¹ [1/s - 1/(s + 6)] + y (0) exp(-6t )

y = 2 u (t - 2) (1 - exp(-6t )) - 2 u (t - 6) (1 - exp(-6t )) + y (0) exp(-6t )