remolatg

## Answered question

2021-02-08

Use Laplace Transform to solve the given equation:
${y}^{″}+6{y}^{\prime }+5y=t\cdot U\left(t-2\right)$
$y\left(0\right)=1$
${y}^{\prime }\left(0\right)=0$

### Answer & Explanation

smallq9

Skilled2021-02-09Added 106 answers

Step 1
Let Laplace transform of y(t) be Y(s). Then Laplace transform of
Also Laplace transform of $u\left(t-2\right)$ is $\frac{{e}^{-2s}}{s}$ Therefore Laplace transform of $tu\left(t-2\right)$ is $\frac{-d}{ds}\frac{{e}^{-2s}}{s}$ or $\frac{\left(2s+1\right){e}^{-2s}}{{s}^{2}}$
Step 2
Finally substituting the results in the equation given:
${s}^{2}Y\left(s\right)-s+6\left(sY\left(s\right)-1\right)+5Y\left(s\right)=\frac{\left(2s+1\right){e}^{-2s}}{{s}^{2}}$
$\left({s}^{2}+6s+5\right)Y\left(s\right)-s-6=\left(2s+1\right)\frac{{e}^{-2s}}{{s}^{2}}$
$\left({s}^{2}+6s+5\right)Y\left(s\right)=\frac{\left(2s+1\right){e}^{-2s}}{{s}^{2}}+s+6$
$Y\left(s\right)=\frac{\left(2s+1\right){e}^{-2s}}{{s}^{2}\left({s}^{2}+6s+5\right)}+\frac{s+6}{{s}^{2}+6s+5}$
$Y\left(s\right)=\frac{\left(2s+1\right){e}^{-2s}}{{s}^{2}\left(s+5\right)\left(s+1\right)}+\frac{s+6}{\left(s+5\right)\left(s+1\right)}$
$Y\left(s\right)={e}^{-2s}\left(-\frac{\frac{32}{25}}{s}+\frac{\frac{7}{5}}{{s}^{2}}+\frac{\frac{5}{4}}{s+1}+\frac{\frac{3}{100}}{s+5}\right)+\frac{\frac{5}{4}}{s+1}-\frac{\frac{1}{4}}{s+5}$
Step 3
Now Laplace inverse of:

Step 4
Thus using time shifting property and solving $Y\left(s\right)={e}^{-2s}\left(-\frac{\frac{32}{25}}{s}+\frac{\frac{7}{5}}{{s}^{2}}+\frac{\frac{5}{4}}{s+1}+\frac{\frac{3}{100}}{s+5}\right)+\frac{\frac{5}{4}}{s+1}-\frac{\frac{1}{4}}{s+5}$:
$y\left(t\right)=\left(-\frac{32}{25}u\left(t-2\right)\right)+\left(\frac{7}{5}\left(t-2\right)u\left(t-2\right)\right)+\left(\frac{5}{4}{e}^{2-t}u\left(t-2\right)\right)+\left(\frac{3}{100}{e}^{10-5t}u\left(t-2\right)\right)+\left(\frac{5}{4}{e}^{-t}u\left(t\right)\right)-\left(\frac{1}{4}{e}^{-5t}u\left(t\right)\right)$

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