Question

Having a problem with solving an IVP using Laplace involving Heavisides, work so far posted in a moment.

Answer 1

Use the Laplace Transform to solve the given Initial Value Problem.
\[y' +y=f(t), \ \ \ y(0)=0 \]where
\[f(t)=
\begin{cases}
0, \ \ \ 0 \leq t<1 \\ 5, \ \ \ t \ge 1
\end{cases}
\]

Answer 2

First, putting the whole thing in the form of a bunch of linear combinations (turning that f(t) into a bunch of heaviside-multiplied functions)

Answer 3

\[f(t)=5\mathcal{U}(t-1)\]

Answer 4

@perl @Concentrationalizing

Answer 5

\[y'+y=5 \ \mathcal{U}(t-1)\]

Answer 6

Taking the laplace of all of that:\[[sY(s)-y(0)]+Y(s)=\frac{5e^{-s}}{s}\]

Answer 7

\[Y(s)(s+1)=\frac{5e^{-s}}{s}\]

Answer 8

\[Y(s)=\frac{5e^{-s}}{s(s+1)}\]

Answer 9

\[\mathcal{L^{-1}}(Y(s))=5 \mathcal{L^{-1}}\left\{\vphantom{} \frac{e^{-s}}{s(s+1)} \right\}\]

Answer 10

By Partial Fraction Decomposition,

Answer 11

\[\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1}\]

Answer 12

By making the identification that,\[a=-1, \ \ \ f(t)=\Bigg[\frac{1}{s}-\frac{1}{s+1}\Bigg]\]We can figure out that our solution is going to be of the form,

Answer 13

\[f(t-1)\mathcal{U}(t-1)\]

Answer 14

\[\mathcal{L^{-1}}\left\{\Bigg[\frac{1}{s}-\frac{1}{s+1}\Bigg]\right\}=1-e^{-s}\]

Answer 15

Multiplying that by the transformed Unit Step and accounting for the shift, we have:\[5 \ \mathcal{U}(t-1)(1-e^{-s-1})\]?

Answer 16

Whoops, those last two variables supposed to be t's, not s's.

Answer 17

(Lol solved my own problem in the process, nvm, don't need help; I forgot about something early on that messed me up before the transform).

Answer 18

And the exponent is supposed to be,

Answer 19

\[e^{-(t-1)}\]

Answer 20

\[Y(s)=5e^{-s}\left( \dfrac{1}{s(s+1)}\right) \implies y(t) = 5\mathcal{U}(t-1)(1-e^{-(t-1)})\]

like this ?

Answer 21

Yeah!

Answer 22

looks good to me