Question: Use the Green function of Problem 6 to solve ${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}{\mathit{a}}^{\mathbf{2}}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,}{\mathit{y}}_{\mathbf{0}}\mathbf{=}{\mathit{y}}_{\mathbf{0}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}\mathbf{.}$

The given expressions are $y^{\text{'}\text{'}}-a^{2}y=e^{-t}$.

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandanti-derivative.

Use the function.

$$\begin{gathered} \begin{array}{c}y=\left\{\begin{array}{l}00<t^{\text{'}}<t\\ {\int }_0^{\infty }\frac{1}{a}\sin a\left(t-t^{\text{'}}\right)0<t^{\text{'}}<t\end{array}\right.\\ y\left(t\right)={\int }_0^{t}d{t}^{\text{'}}\frac{1}{\alpha }\sin a\left(t-t^{\text{'}}\right)e^{-f^{\text{'}}}\end{array} \\ \mathrm{Thus},\mathrm{we}\mathrm{have} \\ \begin{array}{l}\mathrm{I}={\int }_0^{\mathrm{t}}\frac{1}{\mathrm{a}}\sinh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}\\ =\left({\left.\mathrm{uv}\right|}_{\mathrm{t}}^0-{\int }_0^{\mathrm{t}}\mathrm{vdu}\right.\\ =\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\sinh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0-{\int }_0^{\mathrm{t}}\mathrm{acosh}\left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}\right.\\ \mathrm{Again},\mathrm{we}\mathrm{use}\mathrm{integration}\mathrm{by}\mathrm{parts}\mathrm{to}\mathrm{evaluate}\mathrm{the}\mathrm{last}\mathrm{integral},\mathrm{hence}\mathrm{we}\mathrm{have}\\ \begin{array}{l}{\mathrm{I}}_1={\int }_0^{\mathrm{t}}\cosh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}\\ =\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\cosh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0\right.-{\int }_0^{\mathrm{t}}\mathrm{asinh}\left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}\end{array}\\ \mathrm{Thus},\mathrm{we}\mathrm{have}\\ \begin{array}{l}\mathrm{I}=\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\sinh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0-\mathrm{a}{\int }_0^{\mathrm{t}}\cosh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}=\right.\\ \mathrm{I}=\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\sinh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0-\mathrm{a}\left(\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\cosh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0-\mathrm{aI}\right)\right.\right.\end{array}\\ \begin{array}{l}\mathrm{I}=(-0+1\cdot \sinh (\mathrm{at}\left)\right)-\mathrm{a}\left[\left(-{\mathrm{e}}^{\mathrm{t}}\cdot 1+\cosh \left(\mathrm{at}\right)\right)-\mathrm{aI}\right]\\ \mathrm{I}=\sinh \left(\mathrm{at}\right)+{\mathrm{ae}}^{\mathrm{t}}-\mathrm{acosh}\left(\mathrm{at}\right)+{\mathrm{a}}^2\mathrm{I}\end{array}\\ \mathrm{Hence}\mathrm{the}\mathrm{evaluation}\mathrm{of}\mathrm{the}\mathrm{integral}\mathrm{I}\mathrm{is}\mathrm{thus},\\ \begin{array}{l}\mathrm{I}\left(1-{\mathrm{a}}^2\right)=\sinh \left(\mathrm{at}\right)+{\mathrm{ae}}^{\mathrm{t}}-\mathrm{acosh}\left(\mathrm{at}\right)\\ \mathrm{I}=\frac{\sinh \left(\mathrm{at}\right)+{\mathrm{ae}}^{\mathrm{t}}-\mathrm{acosh}\left(\mathrm{at}\right)}{1-{\mathrm{a}}^2}\end{array}\\ \mathrm{And},\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{1}{\mathrm{a}}\mathrm{I}\\ \mathrm{Thus},\mathrm{we}\mathrm{have}\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{\sinh \left(\mathrm{at}\right)+{\mathrm{ae}}^{\mathrm{t}}-\mathrm{acosh}\left(\mathrm{at}\right)}{\mathrm{a}\left(1-{\mathrm{a}}^2\right)}\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{\mathrm{acosh}\left(\mathrm{at}\right)-{\mathrm{ae}}^{\mathrm{t}}-\sinh \left(\mathrm{at}\right)}{\mathrm{a}\left({\mathrm{a}}^2-1\right)}\\ \mathrm{This},\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{for}\mathrm{all}\mathrm{t}>{\mathrm{t}}^{\text{'}}>0\mathrm{else},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{zero}.\end{array} \end{gathered}$$uncaught exception: Invalid chunk

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$$\begin{gathered} \mathrm{Thus},\mathrm{we}\mathrm{have} \\ \begin{array}{l}\mathrm{I}={\int }_0^{\mathrm{t}}\frac{1}{\mathrm{a}}\sinh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}\\ =\left({\left.\mathrm{uv}\right|}_{\mathrm{t}}^0-{\int }_0^{\mathrm{t}}\mathrm{vdu}\right.\\ =\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\sinh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0-{\int }_0^{\mathrm{t}}\mathrm{acosh}\left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}\right.\\ \mathrm{Again},\mathrm{we}\mathrm{use}\mathrm{integration}\mathrm{by}\mathrm{parts}\mathrm{to}\mathrm{evaluate}\mathrm{the}\mathrm{last}\mathrm{integral},\mathrm{hence}\mathrm{we}\mathrm{have}\\ \begin{array}{l}{\mathrm{I}}_1={\int }_0^{\mathrm{t}}\cosh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}\\ =\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\cosh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0\right.-{\int }_0^{\mathrm{t}}\mathrm{asinh}\left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}\end{array}\end{array} \end{gathered}$$

$\begin{array}{l}\mathrm{Thus},\mathrm{we}\mathrm{have}\\ \begin{array}{l}\mathrm{I}=\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\sinh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0-\mathrm{a}{\int }_0^{\mathrm{t}}\cosh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right){\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}{\mathrm{dt}}^{\text{'}}=\right.\\ \mathrm{I}=\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\sinh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0-\mathrm{a}\left(\left(-{\left.{\mathrm{e}}^{-{\mathrm{t}}^{\text{'}}}\cosh \left(\mathrm{a}\left(\mathrm{t}-{\mathrm{t}}^{\text{'}}\right)\right)\right|}_{\mathrm{t}}^0-\mathrm{aI}\right)\right.\right.\end{array}\\ \begin{array}{l}\mathrm{I}=(-0+1\cdot \sinh (\mathrm{at}\left)\right)-\mathrm{a}\left[\left(-{\mathrm{e}}^{\mathrm{t}}\cdot 1+\cosh \left(\mathrm{at}\right)\right)-\mathrm{aI}\right]\\ \mathrm{I}=\sinh \left(\mathrm{at}\right)+{\mathrm{ae}}^{\mathrm{t}}-\mathrm{acosh}\left(\mathrm{at}\right)+{\mathrm{a}}^2\mathrm{I}\end{array}\\ \mathrm{Hence}\mathrm{the}\mathrm{evaluation}\mathrm{of}\mathrm{the}\mathrm{integral}\mathrm{I}\mathrm{is}\mathrm{thus},\\ \begin{array}{l}\mathrm{I}\left(1-{\mathrm{a}}^2\right)=\sinh \left(\mathrm{at}\right)+{\mathrm{ae}}^{\mathrm{t}}-\mathrm{acosh}\left(\mathrm{at}\right)\\ \mathrm{I}=\frac{\sinh \left(\mathrm{at}\right)+{\mathrm{ae}}^{\mathrm{t}}-\mathrm{acosh}\left(\mathrm{at}\right)}{1-{\mathrm{a}}^2}\end{array}\\ \mathrm{And},\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{differential}\mathrm{equation}\mathrm{is}\mathrm{given}\mathrm{by}\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{1}{\mathrm{a}}\mathrm{I}\\ \mathrm{Thus},\mathrm{we}\mathrm{have}\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{\sinh \left(\mathrm{at}\right)+{\mathrm{ae}}^{\mathrm{t}}-\mathrm{acosh}\left(\mathrm{at}\right)}{\mathrm{a}\left(1-{\mathrm{a}}^2\right)}\\ \mathrm{y}\left(\mathrm{t}\right)=\frac{\mathrm{acosh}\left(\mathrm{at}\right)-{\mathrm{ae}}^{\mathrm{t}}-\sinh \left(\mathrm{at}\right)}{\mathrm{a}\left({\mathrm{a}}^2-1\right)}\\ \mathrm{This},\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}\mathrm{for}\mathrm{all}\mathrm{t}>{\mathrm{t}}^{\text{'}}>0\mathrm{else},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{zero}.\end{array}$