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Question Number 142910 by BHOOPENDRA last updated on 07/Jun/21

Answered by qaz last updated on 07/Jun/21

$$\mathrm{y}\left(\mathrm{t}\right)=\mathrm{t}+{\mathrm{e}}^{-2\mathrm{t}}+{\int }_0^{\mathrm{t}}\mathrm{y}\left(\tau \right){\mathrm{e}}^{2(\mathrm{t}-\tau )}\mathrm{d}\tau$$$$\mathcal{L}=\frac{1}{{\mathrm{s}}^2}+\frac{1}{\mathrm{s}+2}+\frac{\mathcal{L}}{\mathrm{s}-2}..............\mathcal{L}=\mathcal{L}\left(\mathrm{y}\left(\mathrm{t}\right)\right)\left(\mathrm{s}\right)$$$$\Rightarrow \mathcal{L}=\frac{({\mathrm{s}}^2+\mathrm{s}+2)(\mathrm{s}-2)}{{\mathrm{s}}^2(\mathrm{s}+2)(\mathrm{s}-3)}$$$$=\frac{2}{3}\cdot \frac{1}{{\mathrm{s}}^2}+\frac{\mathrm{B}}{\mathrm{s}}+\frac{4}{5}\cdot \frac{1}{\mathrm{s}+2}+\frac{14}{45}\cdot \frac{1}{\mathrm{s}-3}$$$$\mathrm{B}=\underset{\mathrm{s}\to 0}{\lim }\frac{\mathrm{d}}{\mathrm{ds}}\left[{\mathrm{s}}^2\mathcal{L}\right]=\underset{\mathrm{s}\to 0}{\lim }\frac{\mathrm{d}}{\mathrm{ds}}\left[\frac{({\mathrm{s}}^2+\mathrm{s}+2)(\mathrm{s}-2)}{(\mathrm{s}+2)(\mathrm{s}-3)}\right]$$$$$$$$=\underset{\mathrm{s}\to 0}{\lim }\frac{[(2\mathrm{s}+1)(\mathrm{s}-2)+({\mathrm{s}}^2+\mathrm{s}+2)](\mathrm{s}+2)(\mathrm{s}-3)-({\mathrm{s}}^2+\mathrm{s}+2)(\mathrm{s}-2)(2\mathrm{s}-1)}{{\left[(\mathrm{s}+2)(\mathrm{s}-3)\right]}^2}$$$$=-\frac{1}{9}$$$$\Rightarrow \mathcal{L}=\frac{2}{3}\cdot \frac{1}{{\mathrm{s}}^2}-\frac{1}{9}\cdot \frac{1}{\mathrm{s}}+\frac{4}{5}\cdot \frac{1}{\mathrm{s}+2}+\frac{14}{45}\cdot \frac{1}{\mathrm{s}-3}$$$$\Rightarrow \mathrm{y}\left(\mathrm{t}\right)={\mathcal{L}}^{-1}=\frac{2}{3}\mathrm{t}-\frac{1}{9}+\frac{4}{5}{\mathrm{e}}^{-2\mathrm{t}}+\frac{14}{45}{\mathrm{e}}^{3\mathrm{t}}$$

Commented by BHOOPENDRA last updated on 07/Jun/21

$$tqsir$$

Answered by Olaf_Thorendsen last updated on 07/Jun/21

$$$$$$y\left(t\right)=t+e^{-2t}+e^{2t}{\int }_0^{t}y\left(\tau \right)e^{-2\tau }d\tau$$$$y\left(0\right)=0+e^0+e^0\times 0=1$$$$e^{-2t}y\left(t\right)=e^{-2t}t+e^{-4t}+{\int }_0^{t}y\left(\tau \right)e^{-2\tau }d\tau$$$$e^{-2t}(y^{\prime }\left(t\right)-2y\left(t\right))=e^{-2t}(1-2t)$$$$-4e^{-4t}+y\left(t\right)e^{-2t}$$$$y^{\prime }\left(t\right)-3y\left(t\right)=1-2t-4e^{-2t}$$$$y\left(t\right)=\frac{2}{3}t-\frac{1}{9}+\frac{4}{5}{e}^{-2t}+{ae}^{3t}$$$$y\left(0\right)=-\frac{1}{9}+\frac{4}{5}+a=1\Rightarrow a=\frac{14}{45}$$$$y\left(t\right)=\frac{2}{3}t-\frac{1}{9}+\frac{4}{5}{e}^{-2t}+\frac{14}{45}{e}^{3t}$$

Commented by BHOOPENDRA last updated on 07/Jun/21

$$tqsir$$

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