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
asked 2020-12-21
Solve the IVP with Laplace Transform:
\(\begin{cases} y"+4y'+4y=(3+t)e^{-2t} \\ y(0)=2 \\ y'(0)=5 \end{cases}\)

Answers (1)

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