Question

Use the Laplace transform to solve the given integral equation or integrodifferential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{sint}\mathbf{-}\underset{\mathbf{0}}{\overset{\mathbf{t}}{\int }}\mathbf{y}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{d\tau }\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The laplace transformation is$y\left(t\right)=\sin t-\frac{1}{2}t\sin t$

Step-by-step solution

Using the transformation formulas:

Firstly using the laplace transformation formula to solve,

$$\begin{gathered} L\left\{1\right\}=\frac{1}{s},L\left\{\sin kt\right\}=\frac{k}{s^2+k^2} \\ \underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau =y\left(t\right)*1 \end{gathered}$$

Using Laplace Transformation formulas;

Take Laplace transformation of both the sides;

Using linearity of the Laplace transform we have;

$$\begin{gathered} L\left\{1\right\}-L\left\{\sin t\right\}-L\left\{\underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau \right\}=L\left\{y\text{'}\right\} \\ L\left\{1\right\}-L\left\{\sin t\right\}-L\left\{\underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau \right\}=sL\left\{y\text{'}\right\}-y\left(0\right) \\ \frac{1}{s}-\frac{1}{s^2+1}-L\left\{\underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau \right\}=sL\left\{y\text{'}\right\}\backslash \mathrm{hfill} \\ \backslash \mathrm{hfill} \\ \frac{1}{s}-\frac{1}{s^2+1}-L\left\{\underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau \right\}=sY\left(s\right) \end{gathered}$$

$$\begin{gathered} L\left\{\underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau \right\}=L\left\{Y\left(t\right)*1\right\} \\ =L\left\{y\left(t\right)\right\}L\left\{1\right\} \\ =Y\left(s\right).\frac{1}{s} \\ Y\left(s\right)=\frac{s^2+1-s}{{\left(s^2+1\right)}^2} \end{gathered}$$

Using the Partial Fraction;

Decompose the fraction using partial fraction technique;

$$\begin{gathered} s^2+1-s=A{s}^3+As+B{s}^2+B+Cs+D \\ {s}^2+1-s=A{s}^3+B{s}^2+(A+C)s+B+D \\ A=0 \\ B=1 \\ A+C=-1 \\ B+D=1 \\ A=0 \\ B=1 \\ C=-1 \\ D=0 \\ \frac{s^2+1-s}{{\left(s^2+1\right)}^2}=\frac{1}{\left(s^2+1\right)}-\frac{s}{{\left(s^2+1\right)}^2} \\ {L}^{-1}\left\{Y\left(s\right)\right\}=L^{-1}\left\{\frac{1}{\left(s^2+1\right)}\right\}-L^{-1}\left\{\frac{2s}{{\left(s^2+1\right)}^2}\right\} \\ =\sin t-\frac{1}{2}t\sin t \\ y\left(t\right)=\sin t-\frac{1}{2}t\sin t \end{gathered}$$

Hence, the final answer is $y\left(t\right)=\sin t-\frac{1}{2}t\mathrm{si}$;