Solving Piece wise ODE using heaviside function laplace transforms

Question

Solve $y''(t)-2y'(t)+y(t)=g(t)$ where $g(t)= \begin{cases}e^{-t} ,& t \in [0,1] \\0 , & t \notin [0,1] \end{cases}$.
Below is my attempt at solving. I get stuck after applying the Heaviside.

---

First, I will define some of the Laplacian transformations.
\begin{align} \mathscr{L}\{y''(t)\}&=s^2 \hat{F}(s)-sy(0)-y'(0) \\ \mathscr{L}\{y'(t)\}&=s \hat{F}(s)-y(0). \\\text{ So, }~~\mathscr{L}\{2y'(t)\}&=2s \hat{F}(s)-2y(0) \\ \mathscr{L}\{y(t)\}&=\hat{F}(s) \\ \mathscr{L}\{g(t)\}&= \mathscr{L}\{e^t-e^tu_1(t)\}, \end{align} where $u_1(t)$ denotes the Heaviside function, activated at $t+1$. And, $$ \mathscr{L}\{e^t-e^tu_1(t)\}=\mathscr{L}\{e^t\}-\mathscr{L}\{e^tu_1(t)\}=\mathscr{L}\{e^t\}-e^{-s}\mathscr{L}\{e^{t+1}\}= \frac{1}{s-1}-e^{-s}\cdot\frac{e}{s-1}$$ via the Heaviside transformation steps.

Then, brought all together, we have...
$$s^2\hat{F}(s)-sy(0)-y'(0)+2s\hat{F}(s)-2y(0)+\hat{F}(s)=\frac{1}{s-1}-e^{-s}\cdot\frac{e}{s-1}. $$

From here, the $\hat{F}(s)$ needs to be isolated to one side of the equation, leaving us with...
$$\hat{F}(s)=\frac{\frac{1+e^{-s+1}}{s-1}+(s+2)y(0)+y'(0)}{(s+1)^2}.$$

And it is here where I am stuck, and I believe that I am missing vital information in terms of the $y(0)$ and $y'(0)$ values or I am not understanding something quite right.

Answer 1

You changed the right side from the task with a factor $e^{-t}$ to the computation where you consistently used $e^t$.

In composing the Laplace transforms of the left side, you missed a sign so that you got $(s+1)^2$ instead of $(s-1)^2$.

In transcribing the Laplace transform of the right side, you missed a sign from $\frac{1}{s-1}-e^{-s}\frac{e}{s-1}$ to the wrong $\frac{1+e^{1-s}}{s-1}$.

---

To verify your solution, use the integrating factor $e^{-t}$ of the left side and set $z=e^{-t}y(t)$. This now satisfies the ODE of the original task $$ z''=e^{-t}(y''-2y'+y)=(u(t)-u(t-1))e^{-2t} $$ Now consider that $f''(t)=u(t)e^{at}$ has a general solution $$f(t)=ct+d+u(t)\frac{e^{at}-1-at}{a^2}$$ to get $$z(t)=ct+d+u(t)\frac{e^{-2t}-1+2t}{4}-e^{-2}u(t-1)\frac{e^{-2(t-1)}-1+2(t-1)}{4}$$

It is quite usual for this type of tasks to demand that also the solution satisfies $y(t)=0$ for $t<0$ so that $c=d=0$ and $$ y(t)=u(t)\frac{e^{-t}+(2t-1)e^t}{4}-u(t-1)\frac{e^{-t}+(2t-3)e^{t-2}}{4}. $$

---

For the modified task as used in the computations you get $$ z''=u(t)-u(t-1)\implies z(t)=cx+d+\frac12u(t)t^2-\frac12u(t-1)(t-1)^2 $$ and with the same assumptions on initial conditions, $$ y(t)=\frac12e^t\bigl[u(t)t^2-u(t-1)(t-1)^2\bigr]. $$

Answer 2

So far your work is correct

You have to do some serious partial fractions to finish the problem.

Answer 3

Starting from Op's answer $$\hat{F}(s)=\frac{1+e^{-s+1}}{(s-1)(s+1)^2}+\frac {y(0)}{(s+1)}+\frac {y(0)+y'(0)}{(s+1)^2}$$ For simplicity susbtitute $A=y(0), B=y'(0)+y(0)$ $$\hat{F}(s)=\frac12({1+e^{-s+1}})\left(\frac 1{(s^2-1)}-\frac 1{(s+1)^2}\right)+\frac {A}{(s+1)}+\frac {B}{(s+1)^2}$$ Take inverse Laplace transform $$f(t)=\frac 12(\sinh(t)-te^{-t})+\frac 12 \mathcal{L^{-1}}(e^{-s+1}\left(\frac 1{(s^2-1)}-\frac 1{(s+1)^2}\right) )+Ae^{-t}+Bte^{-t}$$ $$f(t)=\frac 14e^t+\frac e2 \mathcal{U}(t-1)\left(\sinh(t-1)-(t-1)e^{-t+1} \right )+(A-\frac 14)e^{-t}+(B-\frac 12)te^{-t}$$ Finally $$4f(t)=e^t+ \mathcal{U}(t-1)\left(e^t+(1-2t)e^{-t+2} \right )+(4A-1)e^{-t}+(4B-2)te^{-t}$$ $$4f(t)=e^t+ \mathcal{U}(t-1)\left(e^t+(1-2t)e^{-t+2} \right )+C_1e^{-t}+C_2te^{-t}$$