Use Laplace Transform to solve the given equation: y'' + 6y' + 5y = t cdot U (t-2) y(0) = 1 y' (0) = 0

Question
Laplace transform
Use Laplace Transform to solve the given equation:
$$y'' + 6y' + 5y = t \cdot U (t-2)$$
$$y(0) = 1$$
$$y' (0) = 0$$

2021-02-09
Step 1
Let Laplace transform of y(t) be Y(s). Then Laplace transform of $$y'(t) \text{ and } y"(t) \text{ is } sY(s)-y(0) \text{ or } sY(s)-1 \text{ and } s^2Y(s)-sy(0)-y'(0) \text{ or } s^2Y(s)-s$$
Also Laplace transform of $$u(t-2)$$ is $$\frac{e^{-2s}}{s}$$ Therefore Laplace transform of $$tu(t-2)$$ is $$\frac{-d}{ds}\frac{e^{-2s}}{s}$$ or $$\frac{(2s+1)e^{-2s}}{s^2}$$
Step 2
Finally substituting the results in the equation given:
$$s^2Y(s)-s+6(sY(s)-1)+5Y(s)=\frac{(2s+1)e^{-2s}}{s^2}$$
$$(s^2+6s+5)Y(s)-s-6=(2s+1)\frac {e^{-2s}}{s^2}$$
$$(s^2+6s+5)Y(s)=\frac{(2s+1)e^{-2s}}{s^2}+s+6$$
$$Y(s)=\frac{(2s+1)e^{-2s}}{s^2(s^2+6s+5)}+\frac{s+6}{s^2+6s+5}$$
$$Y(s)=\frac{(2s+1)e^{-2s}}{s^2(s+5)(s+1)}+\frac{s+6}{(s+5)(s+1)}$$
$$Y(s)=e^{-2s}\left(-\frac{\frac{32}{25}}{s}+\frac{\frac{7}{5}}{s^2}+\frac{\frac{5}{4}}{s+1}+\frac{\frac{3}{100}}{s+5}\right)+\frac{\frac{5}{4}}{s+1}-\frac{\frac{1}{4}}{s+5}$$
Step 3
Now Laplace inverse of:
$$\frac{1}{s} \text{ is } u(t)$$
$$\frac{1}{s^2} \text{ is } tu(t)$$
$$\frac{1}{s+1} \text{ is } e^{-t}u(t)$$
$$\frac{1}{s+5} \text{ is } e^{-5t}u(t)$$
Step 4
Thus using time shifting property and solving $$Y(s)=e^{-2s}\left(-\frac{\frac{32}{25}}{s}+\frac{\frac{7}{5}}{s^2}+\frac{\frac{5}{4}}{s+1}+\frac{\frac{3}{100}}{s+5}\right)+\frac{\frac{5}{4}}{s+1}-\frac{\frac{1}{4}}{s+5}$$:
$$y(t)=\left(-\frac{32}{25}u(t-2)\right)+\left(\frac{7}{5}(t-2)u(t-2)\right)+\left(\frac{5}{4}e^{2-t}u(t-2)\right)+\left(\frac{3}{100}e^{10-5t}u(t-2)\right)+\left(\frac{5}{4}e^{-t}u(t)\right)-\left(\frac{1}{4}e^{-5t}u(t)\right)$$

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