Question

Use the Laplace transform and the results of Problem 39 to solve the initial-value problem,

$\begin{array}{l}{\mathbf{y}}^{\mathbf{n}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{sint}\mathbf{+}\mathbf{tsin}\mathbf{;}\\ \mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\end{array}$

Use a graphing utility to graph the solution.

Short answer

The Laplace transformation is$y\left(t\right)=\frac{1}{2}(sint-tcost)+\frac{1}{4}(tsint-t^{2}cost)$

Step-by-step solution

Applying Laplace transformation property

Take Laplace transform of both side of the given equation;

$$\begin{gathered} \mathrm{L}\left\{\mathrm{Sint}\right\}+\mathrm{L}\left\{\mathrm{tsin}\right\}=\mathrm{L}\{\mathrm{y}"\}+\mathrm{I}\left\{\mathrm{y}\right\} \\ ={\mathrm{s}}^2\mathrm{L}\left\{\mathrm{y}\right\}-\mathrm{sy}\left(0\right)-\mathrm{y}\text{'}\left(0\right)+\mathrm{L}\left\{\mathrm{y}\right\} \\ =\left({\mathrm{s}}^2+1\right)\mathrm{L}\left\{\mathrm{y}\right\} \end{gathered}$$

Using Substitution

Denote L{y} = Y(s) and replace in the above equation

$$\begin{gathered} L\left\{\sin t\right\}+L\left\{t\sin t\right\}=\left(s^2+1\right)Y\left(s\right) \\ \frac{1}{s^2+1}+\frac{2s}{{\displaystyle {\left(s^2+1\right)}^2}}=\left(s^2+1\right)Y\left(s\right) \\ Y\left(s\right)=\frac{1}{{\displaystyle {\left(s^2+1\right)}^2}}+\frac{2s}{{\displaystyle {\left(s^2+1\right)}^3}} \end{gathered}$$

Solving further using graph

Using $L^{-1}\left\{\frac{8k^{3}s}{{(s^2+k^2)}^3}\right\}=t\sin kt-k{t}^2\cos k$then it gives when put k=1,

$\begin{array}{c}{L}^{-1}\left\{\frac{8s}{{\left(s^2+1\right)}^3}\right\}=t\sin t-t^2\cos t\\ y\left(t\right)=L^{-1}\left\{\frac{1}{2}\frac{2}{{(s^2+1)}^2}\right\}+L^{-1}\left\{\frac{1}{4}\frac{4.2s}{{(s^2+1)}^3}\right\}\\ =\frac{1}{2}{L}^{-1}\left\{\frac{2}{{(s^2+1)}^2}\right\}+L^{-1}\frac{1}{4}\left\{\frac{8s}{{(s^2+1)}^3}\right\}\\ =\frac{1}{2}(sint-tcost)+\frac{1}{4}(tsint-t^{2}cost)\end{array}$

Hence the final answer is$y\left(t\right)=\frac{1}{2}(sint-tcost)+\frac{1}{4}(tsint-t^{2}cost)$