# Solve the IVP with Laplace Transform: begin{cases} y"+4y'+4y=(3+t)e^{-2t} y(0)=2 y'(0)=5 end{cases}

Question
Laplace transform
Solve the IVP with Laplace Transform:
$$\begin{cases} y"+4y'+4y=(3+t)e^{-2t} \\ y(0)=2 \\ y'(0)=5 \end{cases}$$

2020-12-22
$$\text{Step 1}$$
$$\text{The equation is:}$$
$$y"+4y'+4y=(3+t)e^{-2t}$$
$$\text{Now Laplace equation will be:}$$
$$\left[s^2L\left\{y\right\}-sy(0)-y'(0)\right]+4\left[sL\left\{y\right\}-y(0)\right]+4L\left\{y\right\}=3\frac{1}{s+2}+\frac{1}{(s+2)^2}$$
$$(s^2+4s+4)L\left\{y\right\}-2s-5-8=\frac{3}{s+2}+\frac{1}{(s+2)^2}$$
$$(s^2+4s+4)L\left\{y\right\}-2s-13=\frac{3}{s+2}+\frac{1}{(s+2)^2}$$
$$(s^2+4s+4)L\left\{y\right\}=2s+13+\frac{3}{s+2}+\frac{1}{(s+2)^2}$$
$$\text{Step 2}$$
$$L\left\{y\right\}=\frac{2s^3+21s^2+63s+59}{(s+2)^2(s)}+4s+4$$
$$L\left\{y\right\}=\frac{2s^3+21s^2+63s+59}{(s+2)^4}$$
$$\text{Now take Laplace inverse,}$$
$$y=L^{-1}\left\{\frac{2s^3+21s^2+63s+59}{(s+2)^4}\right\}$$
$$\text{By the formula of Laplace transformation the final solution will be:}$$
$$y=(2+9x)e^{-2x}$$

### Relevant Questions

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Solve the following IVP using Laplace Transform
$$y′′+3y′+2y=e^(-t), y(0)=0 y′(0)=0$$
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$$y(0)=0$$
$$y'(0)=1$$
$${y}\text{}-{4}{y}={e}^{{-{3}{t}}},{y}{\left({0}\right)}={0},{y}'{\left({0}\right)}={2}$$
a) $$\frac{14}{{20}}{e}^{{{2}{t}}}-\frac{5}{{30}}{e}^{{-{2}{t}}}-\frac{9}{{30}}{e}^{{-{6}{t}}}$$
b) $$\frac{11}{{20}}{e}^{{{2}{t}}}-\frac{51}{{20}}{e}^{{-{2}{t}}}-\frac{4}{{20}}{e}^{{-{3}{t}}}$$
c) $$\frac{14}{{15}}{e}^{{{2}{t}}}-\frac{5}{{10}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$
d) $$\frac{14}{{20}}{e}^{{{2}{t}}}+\frac{5}{{20}}{e}^{{-{2}{t}}}-\frac{9}{{20}}{e}^{{-{3}{t}}}$$