Question

Find the eigen value and eigen function of homogeneous integral equation

y(x)= 𝜆$\int$K(x,t)y(t)dt where K(x,t)= sinx sin(t-1) -𝜋<=x<=t

sint sin(x-1) t<=x<= 𝜋

Answer 1

Answer:-

This question is incomplete but we can take $y(x)=\lambda \int_{0}^{1} e^{x+1} y(t) d t$ to understand the problem.

The given equation can be written as

$y(x)=\lambda e^{x} \int_{0}^{1} e^{t} y(t) d t \ldots(\mathrm{i})$

Assume that $\quad C=\int_{0}^{1} e^{t} y(t) d t \ldots(\mathrm{ii})$

Then, from Eq. (i), we have $u(x)=\lambda C e^{x} \ldots . . (iii)$

$\Rightarrow y(t)=\lambda C e^{t}$

Putting this value in eq. (ii), we get

$\begin{aligned} &C=\int_{0}^{1} e^{t}\left(\lambda C e^{t}\right) d t=\lambda C\left[\frac{e^{2 t}}{2}\right]_{0}^{1} \\ &\Rightarrow C=\frac{\lambda C}{2}\left(e^{2}-1\right) \Rightarrow C\left[1-\frac{\lambda}{2}\right]\left(e^{2}-1\right)=0 \end{aligned}$

If $\mathrm{C}=0$ , then Eq. (iii) gives ${y}({x})=0$ .

We, therefore assume that for non-zero solution of Eq. (i).

If $C \neq 0$ , then Eq. (iii) gives

$1-\frac{\lambda}{2}\left(e^{2}-1\right)=0$

$\Rightarrow \lambda=\frac{2}{e^{2}-1}$

Which is an eigen value of Eq. (i).

To find the corresponding eigen function, putting the value of $\lambda$ in Eq. (iii), we get

$y(x)=\frac{2 C}{e^{2}-1} \cdot e^{x}$

The eigen function is $\mathrm{e}^{{x}}.$