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Tried solving those for the past few days and could not get far enough to find the answers anyone can help with those?

Using the Laplace transform, find the the solution $y(t)$ of the ODE

$$2y''(t)+20y'(t)+32y(t)=2u(t)$$

where $y'(0)=0$, $y(0)=1$, and $u(t)$ is the unit step function.

Expected solution: $$y(t) = 0.0625 + 1.25e^{-2t} - 0.3125e^{-8t}.$$

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  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Commented Sep 13, 2022 at 15:11
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    $\begingroup$ The procedure is as follows: compute the Laplace transform of both sides of the equation, solve for $Y(s)$, express $Y(s)$ is terms of simple fractions, use the inverse Laplace transform to get the solution $y(t)$. $\endgroup$
    – KBS
    Commented Sep 13, 2022 at 15:22
  • $\begingroup$ It feels like I don't have the math level to find the answer. I'm having trouble solving for Y(s). I might need to go back to calculus before doing this but thanks for the reply :) $\endgroup$
    – Caio
    Commented Sep 13, 2022 at 15:37
  • $\begingroup$ You have tables for the Laplace transform and its inverse. See e.g. en.wikipedia.org/wiki/Laplace_transform $\endgroup$
    – KBS
    Commented Sep 13, 2022 at 15:41
  • $\begingroup$ i'll try this ig $\endgroup$
    – Max0815
    Commented Sep 13, 2022 at 17:10

1 Answer 1

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We want to solve the IVP $$2y''(t)+20y'(t)+32y(t)=2H(t)$$ with conditions $y'(0)=0$ and $y(0)=1$ using Laplace transformations.

Begin by removing the factor of $2$ and taking the Laplace transform to get \begin{align*} \mathcal L\{y''(t)\}(s)+10\mathcal L\{y'(t)\}(s)+16\mathcal L\{y(t)\}(s)&=\mathcal L\{H(t)\}(s)\\ \left[s^2Y(s)-sy(0)-y'(0)\right]+10\left[sY(s)-y(0)\right]+16[Y(s)]&=\frac1s\\ s^2Y(s)-s+10sY(s)-10+16Y(s)&=\frac1s\\ \left(s^2+10s+16\right)Y(s)&=\frac1s+s+10\\ Y(s)&=\frac{s^2+10s+1}{s(s+8)(s+2)} \end{align*} Upon this result we can perform PFD to get $$Y(s)=\frac{s^2+10s+1}{s(s+8)(s+2)}=\frac1{16s}+\frac5{4(2+s)}-\frac5{16(8+s)}$$ Now, take the inverse Laplace transform to get \begin{align*} \mathcal{L}^{-1}\{Y(s)\}&=\frac1{16}\mathcal{L}^{-1}\left\{\frac1{s}\right\}+\frac{5}{4}\mathcal{L}^{-1}\left\{\frac1{2+s}\right\}-\frac5{16}\mathcal{L}^{-1}\left\{\frac1{8+s}\right\}\\ y(t)&=\boxed{\frac1{16}+\frac5{4}e^{-2t}-\frac{5}{16}e^{-8t}} \end{align*}

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    $\begingroup$ I think you have a couple of errors going from the second to the third line of your solution. $\endgroup$ Commented Sep 13, 2022 at 17:45
  • $\begingroup$ @JohnBarber ok its now fixed lol was in class when i wrote this so did not pay too much attention oops $\endgroup$
    – Max0815
    Commented Sep 13, 2022 at 17:55

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