Q15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given integral equation or integro-differential equation for y(t) $\mathit{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{3}\underset{0}{\overset{t}{\int }}y\left(v\right)sin(t-v)dv\mathbf{=}\mathit{t}$       

Step-by-Step Solution

Answer

Integral equation or integro-differential equation for y(t) is .$\frac{1}{4}t+\frac{3}{8}\sin 2t$

Step 1: Given the integral equation is,

$\begin{array}{c}y\left(t\right)+3{\int }_0^{t}y\left(v\right)\sin (t-v)dv=t\\ y\left(t\right)+3\left[y\right(t)\star \sin t]=t\end{array}$                             [Using      $g\star h={\int }_0^{t}g(t-v)h\left(v\right)dv$

Step 2: Taking Laplace transform on both sides,

$\begin{array}{c}y\left(s\right)+3\left[y\left(s\right)\cdot \frac{1}{s^2+1}\right]=\frac{1}{s^2}\\ [1+\frac{3}{s^2+1}]y\left(s\right)=\frac{1}{s^2}\\ y\left(s\right)=\frac{s^2+1}{s^2(s^2+4)}\dots \dots \left(1\right)\end{array}$

Step 3: Using the partial function, we get

$\frac{s^2+1}{s^2(s^2+4)}=\frac{1}{4s^2}+\frac{3}{4(s^2+4)}$

Now, equation (1) becomes

$\begin{array}{c}y\left(s\right)=\frac{1}{4s^2}+\frac{3}{4(s^2+4)}\\ =\frac{1}{4s^2}+\frac{3}{8}\frac{2}{(s^2+4)}\end{array}$

Step 4: Taking inverse Laplace transform, we get

$y\left(t\right)=\frac{1}{4}t+\frac{3}{8}\sin 2t$