 # 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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