Question

In Problems 35–42 use the Laplace transform to solve the given

equation.

$\mathbf{41}\mathbf{.}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{t}\mathbf{+}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}\mathbf{y}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{\tau }\mathbf{)}\mathbf{d}\mathbf{\tau }\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The equation is$y\left(t\right)=t+1+\frac{1}{2!}{t}^2$

Step-by-step solution

To Find the laplace transform equation.

Take Laplace transform of both side of the given equation

$L\left\{cost\right\}+L\underset{0}{\overset{t}{\int }}y\left(\tau \right)cos(t-\tau )d\tau =L\left\{y\text{'}\right\}$

$L\left\{cost\right\}+L\underset{0}{\overset{t}{\int }}y\left(\tau \right)cos(t-\tau )d\tau \stackrel{*}{=}sL\left\{y\right\}-y\left(0\right)$

$\frac{s}{s^2+1}+LL\underset{0}{\overset{t}{\int }}y\left(\tau \right)cos(t-\tau )d\tau =sL\left\{y\right\}-1y\left(0\right)=1,L\left\{coskt\right\}=\frac{s}{s^2+k^2}$

Denote and replace in the above equation.

$\frac{s}{s^2+1}+L{\int }_0^{t}y\left(\tau \right)cos(t-\tau )d\tau =sY\left(s\right)-1.......\left(1\right)$

Find the equation.

If functions f and g are piecewise continuous on the interval, $[0,\infty ),$then

$f*g=\underset{0}{\overset{t}{\int }}f\left(\tau \right)g(t-\tau )d\tau$

Use the above formula and Laplace transform, to find $L{\int }_0^{t}y\left(\tau \right)cos(t-\tau )d\tau$

$\begin{array}{rcl}L{\int }_0^{t}y\left(\tau \right)cos(t-\tau )d\tau & =& L\left\{y\right(t)*cost\}=L\left\{y\right(t\left)\right\}L\left\{cost\right\}\left(L\right\{f*g\}=L\{f\left\}L\right\{g\left\}\right)=Y\left(s\right)\times \frac{s}{s^2+1}\\ L\left\{y\right(t\left)\right\}& =& Y\left(s\right),L\left\{coskt\right\}=\frac{s}{s^2+k^2}\end{array}$

Substitute in (1), to get.

$\begin{array}{rcl}\frac{s}{s^2+1}+Y\left(s\right)\frac{s}{s^2+1}& =& sY\left(s\right)-1\\ s+sY\left(s\right)& =& s^{3}Y\left(s\right)+sY\left(s\right)-s^2-1\\ s^{3}Y\left(s\right)& =& s+s^2+1\\ Y\left(s\right)& =& \frac{1}{s^2}+\frac{1}{s}+\frac{1}{s^3}\end{array}$

Final proof

Take inverse Laplace transform of both side

$L^{-1}\left\{Y\right(s\left)\right\}=L^{-1}\frac{1}{s^2}+L^{-1}\frac{1}{s}+\frac{1}{2!}{L}^{-1}\frac{2!}{s^3}\stackrel{\left(**\right)}{=}t+1+\frac{1}{2!}{t}^2{L}^{-1}\frac{n!}{s^{n+1}}=t^n,L^{-1}\frac{1}{s}=1(**)$

So,

$y\left(t\right)=t+1+\frac{1}{2!}{t}^2$