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Section 6.2 – Solution of Initial Value Problems [via Laplace Transforms]

Directions. The following are review problems for the section. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it. 
  1. Use the Laplace transform to solve the given initial value problem
    1. \(y''+3y'+2y=4t,\qquad\) \(y(0)=1,\quad\)  \(y'(0)=0\)
    2. \(y''+9y=\cos 2t,\qquad\) \(y(0)=0,\quad\) \(y'(0)=1\)
    3. \(y''-2y'+2y=e^{-t},\qquad\) \( y(0)=0,\quad\) \( y'(0)=1\)​

      1. \(y(t)=\mathcal{L}^{-1}\{Y(s)\}=-3+2t+6e^{-t}-2e^{-2t}\)
      2. \(y(t)=\mathcal{L}^{-1}\{Y(s)\}=\dfrac{1}{5}\cos(2t)-\dfrac{1}{5}\cos(3t)+\dfrac{1}{3}\sin(3t)\)
      3. \(y(t)=\mathcal{L}^{-1}\{Y(s)\}=\dfrac{1}{5}e^{-t}-\dfrac{1}{5}e^t\cos(t)+\dfrac{7}{5}e^t\sin(t)\)
      Video Errata: In part (c), at the 46:58 mark, we should have that \(C=\dfrac{8}{5}\). We then have that \(Y(s)=\dfrac{A}{s+1}+\dfrac{Bs+C}{s^2-2s+2}=\dfrac{1/5}{s+1}+\dfrac{1}{5}\cdot \dfrac{-s+8}{(s-1)^2+1}\). Moving forward, we have \(Y(s)=\dfrac{1}{5}\cdot\dfrac{1}{s+1}-\dfrac{1}{5}\cdot\dfrac{s-1}{(s-1)^2+1}+\dfrac{7}{5}\cdot\dfrac{1}{(s-1)^2+1}\). Our final answer is then \(y(t)=\mathcal{L}^{-1}\{Y(s)\}=\dfrac{1}{5}e^{-t}-\dfrac{1}{5}e^t\cos(t)+\dfrac{7}{5}e^t\sin(t)\). A note of this correction is made at the 57:39 mark.


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      MATH 308 WIR22A V52