Question

In Problems 35–42 use the Laplace transform to solve the given

Equation.

36.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathbf{y}\mathbf{=}\mathbf{t}{\mathbf{e}}^{\mathbf{t}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$y\left(t\right)=\frac{6}{169}{e}^t+\frac{1}{13}t{e}^t-\frac{6}{169}{e}^{4t}\cos 2t+\frac{5}{338}{e}^{4t}\sin 2t$

Step-by-step solution

Given Information.

$y\text{'}\text{'}-8y\text{'}+20y=t{e}^t,y\left(0\right)=0,y\text{'}\left(0\right)=0$

Apply Laplace Transform

Take Laplace transform of both side of the given equation

$L\left\{y\text{'}\text{'}-8y\text{'}+20y\right\}=L\left\{t{e}^t\right\}$

$L\left\{f^n\left(t\right)\right\}=s^{n}L\left\{f\right(t\left)\right\}-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)\dots .f^{n-1}\left(0\right)andL\left\{e^{at}f\left(t\right)\right\}={\left.L\left\{f\right(t\left)\right\}\right|}_{s\to s-a}$

To replace $L\left\{y\right\}=Y\left(s\right)$of the equation

$\left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]-8\left[sY\right(s)-y(0\left)\right]+20Y\left(s\right)={\left.\frac{1}{s^2}\right|}_{s\to s-1}$

$\left(s^2-8s+20\right)Y\left(s\right)=\frac{1}{{(s-1)}^2}$

$Y\left(s\right)=\frac{1}{{(s-1)}^2\left(s^2-8s+20\right)}$

$Y\left(s\right)=\frac{1}{{(s-1)}^2\left({(s-4)}^2+2^2\right)}$

Apply Partial fraction

$\frac{1}{{(s-1)}^2\left({(s-4)}^2+2^2\right)}=\frac{A}{s-1}+\frac{B}{{(s-1)}^2}+\frac{Cs+D}{{(s-4)}^2+2^2}$

Equating the coefficients

$A=\frac{6}{169}B=\frac{1}{13}C=-\frac{6}{169}D=\frac{29}{169}$

$Y\left(s\right)=\frac{6}{169}\frac{1}{s-1}+\frac{1}{13}\frac{1}{{(s-1)}^2}-\frac{6}{169}\frac{s}{{(s-4)}^2+2^2}+\frac{29}{169}\frac{1}{{(s-4)}^2+2^2}$

Apply Inverse Laplace Transform

$L^{-1}\left\{Y\right(s\left)\right\}=L^{-1}\left\{\frac{6}{169}\frac{1}{s-1}+\frac{1}{13}\frac{1}{{(s-1)}^2}-\frac{6}{169}\frac{s}{{(s-4)}^2+2^2}+\frac{29}{169}\frac{1}{{(s-4)}^2+2^2}\right\}$

$y\left(t\right)=\frac{6}{169}{L}^{-1}\left\{\frac{1}{s-1}\right\}+\frac{1}{13}{L}^{-1}\left\{\frac{1}{{(s-1)}^2}\right\}-\frac{6}{169}{L}^{-1}\left\{\frac{s}{{(s-4)}^2+2^2}\right\}+\frac{5}{338}{L}^{-1}\left\{\frac{2}{{(s-4)}^2+2^2}\right\}$

By using this formula $L^{-1}\left\{F\right(s-a\left)\right\}=e^{at}f\left(t\right)$

$y\left(t\right)=\frac{6}{169}{e}^t+\frac{1}{13}t{e}^t-\frac{6}{169}{e}^{4t}\cos 2t+\frac{5}{338}{e}^{4t}\sin 2t$