Question

In Problems 9-14use the Laplace transform to solve the given initialvalue problem. Use the table of Laplace transforms in Appendix C as needed.

$\mathbf{10}\mathbf{.}\mathbf{}{\mathbf{y}}^{\mathbf{t}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{te}}^{\mathbf{t}}\mathbf{sint}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The Laplace transfom of${\mathrm{y}}^{\mathrm{t}}-\mathrm{y}={\mathrm{te}}^{\mathrm{t}}\mathrm{sint}\mathrm{is}\mathrm{y}={\mathrm{e}}^{\mathrm{t}}(\mathrm{sint}-\mathrm{tcost})$

Step-by-step solution

Determine the derivatives of transforms

The Laplace transform converts a linear differential equation to an algebraic equation, which may subsequently be solved using algebraic principles.

The inverse Laplace transform can then be used to solve the original differential equation.

$F(s={\int }_0^{\infty }f(t)e^{-st}t\text{'}$

Determine the laplace transforms in appendix

We have to solve the given initial value problem using Laplace transform

${\mathrm{y}}^{\text{'}}-\mathrm{y}={\mathrm{te}}^{\mathrm{t}}\mathrm{sint}$

Performing Laplace transform of the above equation, we get

$\mathrm{L}\{{\mathrm{y}}^{\text{'}}-\mathrm{y}\}=\mathrm{L}\left\{{\mathrm{te}}^{\mathrm{t}}\mathrm{sint}\right\}$

Using Appendix III for the required Laplace Transforms, we have

$\mathrm{L}\left\{{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}\right\}=\frac{1}{{(\mathrm{s}-1)}^2+1}$

We know that

$\mathrm{L}\left\{{\mathrm{t}}^{\mathrm{n}}\mathrm{f}\left(\mathrm{t}\right)\right\}={(-1)}^{\mathrm{n}}\frac{{\mathrm{d}}^{\mathrm{n}}}{{\mathrm{ds}}^{\mathrm{n}}}\left[\mathrm{F}\right(\mathrm{s}\left)\right]$

therefore, we have

$\mathrm{L}\left\{{\mathrm{te}}^{\mathrm{t}}\mathrm{sint}\right\}=(-1)\frac{\mathrm{d}}{\mathrm{ds}}\left[\frac{1}{{(\mathrm{s}-1)}^2+1}\right]$

$\mathrm{L}\left\{{\mathrm{te}}^{\mathrm{t}}\mathrm{sint}\right\}=\frac{2(\mathrm{s}-1)}{{[{(\mathrm{s}-1)}^2+1]}^2}$

Hence, we have the Laplace Transformed equation as

$\mathrm{L}\{{\mathrm{y}}^{\text{'}}-\mathrm{y}\}=\mathrm{L}\left\{{\mathrm{te}}^{\mathrm{t}}\mathrm{sint}\right\}$

$\mathrm{sL}\left\{\mathrm{y}\right\}-\mathrm{y}\left(0\right)-\mathrm{L}\left\{\mathrm{y}\right\}=\frac{2(\mathrm{s}-1)}{{[{(\mathrm{s}-1)}^2+1]}^2}$

$\mathrm{sL}\left\{\mathrm{y}\right\}-\mathrm{L}\left\{\mathrm{y}\right\}=\frac{2(\mathrm{s}-1)}{{[{(\mathrm{s}-1)}^2+1]}^2}$

$(\mathrm{s}-1)\mathrm{L}\left\{\mathrm{y}\right\}=\frac{2(\mathrm{s}-1)}{{[{(\mathrm{s}-1)}^2+1]}^2}$

$\mathrm{L}\left\{\mathrm{y}\right\}=\frac{2}{{[{(\mathrm{s}-1)}^2+1]}^2}$

By partial fractions, we have

We know that

$\mathrm{L}\left\{{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}\right\}=\frac{1}{{(\mathrm{s}-1)}^2+1}$

Hence, we have

${\mathrm{L}}^{-1}\left\{\frac{1}{{(\mathrm{s}-1)}^2+1}\right\}={\mathrm{e}}^{\mathrm{t}}\mathrm{sint}$

Similarly, we know that

$\mathrm{L}\left\{{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}\right\}=\frac{(\mathrm{s}-1)}{{(\mathrm{s}-1)}^2+1}$

Therefore,

$\mathrm{L}\left\{{\mathrm{te}}^{\mathrm{t}}\mathrm{cost}\right\}=(-1)\frac{\mathrm{d}}{\mathrm{ds}}\left[\mathrm{L}\left\{{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}\right\}\right]$

$\mathrm{L}\left\{{\mathrm{te}}^{\mathrm{t}}\mathrm{cost}\right\}=(-1)\frac{\mathrm{d}}{\mathrm{ds}}\left[\frac{(\mathrm{s}-1)}{{(\mathrm{s}-1)}^2+1}\right]$

$\mathrm{L}\left\{{\mathrm{te}}^{\mathrm{t}}\mathrm{cost}\right\}=\frac{\mathrm{s}(\mathrm{s}-2)}{{[{(\mathrm{s}-1)}^2+1]}^2}$

Hece, we have

${\mathrm{L}}^{-1}\left\{\frac{\mathrm{s}(\mathrm{s}-2)}{{[{(\mathrm{s}-1)}^2+1]}^2}\right\}={\mathrm{te}}^{\mathrm{t}}\mathrm{cost}$

Now, by performing Inverse Laplace Transform, we have

$\mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{[{(\mathrm{s}-1)}^2+1]}-\frac{\mathrm{s}(\mathrm{s}-2)}{{[{(\mathrm{s}-1)}^2+1]}^2}$

$\mathrm{y}={\mathrm{L}}^{-1}\left\{\frac{1}{[{(\mathrm{s}-1)}^2+1]}\right\}-{\mathrm{L}}^{-1}\left\{\frac{\mathrm{s}(\mathrm{s}-2)}{{[{(\mathrm{s}-1)}^2+1]}^2}\right\}$

$y=e^t\sin t-t{e}^t\cos t$

$y=e^t(\sin t-t\cos t)$

Therfore, the Laplace transfom of $y^t-y=t{e}^t\sin t$is $y=e^t(\sin t-t\cos t)$.