Question 1

Problem 1: Use the Laplace Transforms to solve:

$y'' - y' = {e}^( - 3t) , \\ y(0) =y'(0) =0$

$Problem 1: Use the Laplace Transforms to solve: y'' - y' = {e}^( - 3t) , \\ y(0) =y-example-1$

Question 2

Transforming the ODE yields

$L\left\{y'' - y'\right\} = L\left\{e^(-3t)\right\}$

$(s^2 Y(s) - sy(0) - y'(0)) - (s Y(s) - y(0)) = \frac1{s+3}$

$(s^2 - s) Y(s) = \frac1{s+3}$

$Y(s) = \frac1{(s^2 - s)(s+3)} = \frac1{s(s-1)(s+3)}$

Partial fractions:

$Y(s) = \frac as + \frac b{s-1} + \frac c{s+3}$

$\implies 1 = a(s-1)(s+3) + b s(s+3) + c s(s - 1)$

$\implies 1 = -3 a + (2 a + 3 b - c) s + (a + b + c) s^2$

$\implies \begin{cases}-3a=1 \\ 2a+3b-c=0 \\ a+b+c=0\end{cases} \implies a=-\frac13, b=\frac14, c=\frac1{12}$

$\implies Y(s) = -\frac13 * \frac1s + \frac14 * \frac1{s-1} + \frac1{12} * \frac1{s+3}$

Take the inverse transform:

$\boxed{y(t) = -\frac13 + \frac{e^t}4 + (e^(-3t))/(12)}$

Please log in or register to add a comment.

Please log in or register to answer this question.