Question

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

Equation.

37.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{t}\mathbf{-}\mathbf{t}\mathbf{u}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$y\left(t\right)=-\frac{6}{25}+\frac{1}{5}{t}^2+\frac{3}{2}{e}^{-t}-\frac{13}{50}{e}^{-5t}+\left[-\frac{4}{25}-\frac{1}{5}{(t-2)}^2+\frac{1}{4}{e}^{-(t-2)}-\frac{9}{100}{e}^{-5(t-2)}\right]u(t-2)$

Step-by-step solution

Given Information.

$y\text{'}\text{'}+6y\text{'}+5y=t-tu(t-2),y\left(0\right)=1,y\text{'}\left(0\right)=0$

Apply Laplace Transform

Take Laplace transform of both side of the given equation

$L\left\{y\text{'}\text{'}+6y\text{'}+5y\right\}=L\{t-tu(t-2\left)\right\}$

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

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

$\left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+6\left[sY\right(s)-y(0\left)\right]+5Y\left(s\right)=\frac{1}{s^2}-\frac{e^{-2s}}{s^2}-\frac{2e^{-2s}}{s}$

$\left(s^2+6s+5\right)Y\left(s\right)=\frac{1}{s^2}-\frac{e^{-2s}}{s^2}-\frac{2e^{-2s}}{s}+s+6$

$Y\left(s\right)=\frac{s^3+6s^2+1}{s^2(s+1)(s+5)}-\frac{e^{-2s}}{s^2(s+1)(s+5)}-\frac{2e^{-2s}}{s(s+1)(s+5)}$

Apply Inverse Laplace Transform

$Y\left(s\right)=\left[-\frac{6}{25}\frac{1}{s}+\frac{1}{5}\frac{1}{s^2}+\frac{3}{2}\frac{1}{s+1}-\frac{13}{50}\frac{1}{s+5}\right]-\left[-\frac{6}{25}\frac{1}{s}+\frac{1}{5}\frac{1}{s^2}+\frac{1}{4}\frac{1}{s+1}-\frac{1}{100}\frac{1}{s+5}\right]$

$e^{-2s}-\left[\frac{2}{5}\frac{1}{5}-\frac{1}{2}\frac{1}{s+1}+\frac{1}{10}\frac{1}{s+5}\right]e^{-2s}$

$y\left(t\right)=-\frac{6}{25}+\frac{1}{5}{t}^2+\frac{3}{2}{e}^{-t}-\frac{13}{50}{e}^{-5t}+\left[-\frac{4}{25}-\frac{1}{5}{(t-2)}^2+\frac{1}{4}{e}^{-(t-2)}-\frac{9}{100}{e}^{-5(t-2)}\right]u(t-2)$