Question

Use the Laplace transform to solve the given initial-value problem.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\sqrt{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\sqrt{\mathbf{2}\mathbf{t}}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{10}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

The Laplace transform of the given derivative equation

$Y\left(t\right)=-\sqrt{2}\sin \sqrt{2t}+2\sin t+10\cos t$

Step-by-step solution

Theorem of Transform of a Derivative.

If $\mathit{f}\mathbf{,}{\mathit{f}}^{\mathbf{\text{'}}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{f}}^{\left(n-1\right)}$are continuous on role="math" localid="1663969400282" $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$ and are of exponential order and if ${\mathit{f}}^{\left(n\right)}\left(t\right)$is a piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$, then

$\mathit{L}\left\{f^{\left(n\right)}\left(t\right)\right\}\mathbf{=}{\mathit{s}}^{\mathbf{n}}\mathit{F}\left(s\right)\mathbf{-}{\mathit{s}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathit{f}\left(0\right)\mathbf{-}{\mathit{s}}^{\mathbf{n}\mathbf{-}\mathbf{2}}{\mathit{f}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{-}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{-}{\mathit{f}}^{\left(n-1\right)}\left(0\right)$

Where $\mathit{F}\left(s\right)\mathbf{=}\mathit{L}\left\{f\left(t\right)\right\}$

The given derivative equation is $y^{\text{'}\text{'}}+y=\sqrt{2}\sin \sqrt{2t}$with the initial condition as

$y\left(0\right)=10,y^{\text{'}}\left(0\right)=0$

Take the Laplace Transform on both the sides of the equation,

$\begin{array}{c}L\left\{y^{\text{'}\text{'}}\right\}+L\left\{y\right\}=\sqrt{2}L\left\{\sin \sqrt{2}t\right\}\\ \left[s^{2}Y\left(s\right)-s\left(10\right)-s^0\left(0\right)\right]+Y\left(s\right)=\sqrt{2}\times \frac{\sqrt{2}}{s^2+2}\\ Y\left(s\right)\left(s^2+1\right)-10s=\frac{2}{s^2+2}\\ \end{array}$

Using partial fraction.

Now solve for $Y\left(s\right)$,

$Y\left(s\right)=\frac{2}{\left(s^2+2\right)\left(s^2+1\right)}+\frac{10s}{s^2+1}$

Use partial fractions on the first term,

$\begin{array}{c}\frac{2}{\left(s^2+2\right)\left(s^2+1\right)}=\frac{As+B}{s^2+2}+\frac{Cs+D}{s^2+1}\\ \left(As+B\right)\left(s^2+1\right)+\left(s^2+2\right)\left(Cs+D\right)=2\\ A{s}^3+As+B{s}^2+B+C{s}^3+D{s}^2+2Cs+2D=2\end{array}$

The system of equation generated and the values found by partial fractions,

$\begin{array}{c}A+C=0\\ B+D=0\\ A+2C=0\\ B+2D=2\\ A=0\\ B=-2\\ C=0\\ D=2\end{array}$

Substitution of the partial fraction values.

Substitute the values found by the partial fraction,

$Y\left(s\right)=\frac{-2}{s^2+2}+\frac{2}{s^2+1}+\frac{10s}{s^2+1}$

Manipulate the first term,

$\begin{array}{c}Y\left(s\right)=\frac{-2}{\sqrt{2}}\times \frac{-2\times \frac{\sqrt{2}}{-2}}{s^2+2}+\frac{2}{s^2+1}+\frac{10s}{s^2+1}\\ Y\left(s\right)=\frac{-2}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\times \frac{-2\times \frac{\sqrt{2}}{-2}}{s^2+2}+\frac{2}{s^2+1}+\frac{10s}{s^2+1}\\ Y\left(s\right)=-\sqrt{2}\times \frac{\sqrt{2}}{s^2+2}+\frac{2}{s^2+1}+\frac{10s}{s^2+1}\end{array}$

Finding the inverse Laplace Transform.

Now take the inverse Laplace Transform of both the sides of the equation,

$\begin{array}{c}{L}^{-1}\left\{Y\left(s\right)\right\}=-\sqrt{2}{L}^{-1}\frac{\sqrt{2}}{s^2+2}+2L^{-1}\frac{1}{s^2+1}+10L^{-1}\frac{s}{s^2+1}\\ y\left(t\right)=-\sqrt{2}\sin \sqrt{2}t+2\sin t+10\cos t\end{array}$

Thus the Laplace Transform of the given derivative equation is

$Y\left(t\right)=-\sqrt{2}\sin \sqrt{2t}+2\sin t+10\cos t$