Finding k values for non-zero solutions in integral equations?

[Alone]

Hi, I have the next question from Pipkin's textbook on Integral Equations:

question 9, page 10:
Find the values of $$k$$ for which the following equation has solutions that aren't identically zero. If $$k\neq 0$$, find representative solutions:$$\int_{-\pi}^{\pi} \sin(x+y)u(y)dy = ku(x)$$

What I have done so far is the following:denote by $$c_1=\int \cos(y)u(y)dy,\ c_2=\int \sin(y)u(y)dy$$
so we have:$$c_1\sin(x)+c_2 \cos(x)=ku(x)$$, so I found represntaive solutions, but how do I find the values of k?Thanks in advance.

 

[alyafey22]

I think something is missing maybe the value of u(x) at some points ?

 

[I like Serena]

Hi, I have the next question from Pipkin's textbook on Integral Equations:

question 9, page 10:
Find the values of $$k$$ for which the following equation has solutions that aren't identically zero. If $$k\neq 0$$, find representative solutions:$$\int_{-\pi}^{\pi} \sin(x+y)u(y)dy = ku(x)$$

What I have done so far is the following:denote by $$c_1=\int \cos(y)u(y)dy,\ c_2=\int \sin(y)u(y)dy$$
so we have:$$c_1\sin(x)+c_2 \cos(x)=ku(x)$$, so I found represntaive solutions, but how do I find the values of k?Thanks in advance.

Hi Alan! :)

You have $$c_1\sin(x)+c_2 \cos(x)=ku(x)$$
So $$u(x) = \frac 1 k (c_1\sin(x)+c_2 \cos(x))$$
Have you tried to substitute that into your original equation?

 

[Alone]

Hi Alan! :)

You have $$c_1\sin(x)+c_2 \cos(x)=ku(x)$$
So $$u(x) = \frac 1 k (c_1\sin(x)+c_2 \cos(x))$$
Have you tried to substitute that into your original equation?

And then what do I get? I get the following:

$$\int_{-\pi}^{\pi}\sin(x+y)(c_1\sin y+c_2\cos y)/k dy = c_1\sin x + c_2 \cos x = ((c_1\cos x )\pi+(c_2\sin x)\pi)/k$$

How to procceed from there?

Edit:
We should have: $c_1 \pi/k = c_2 , c_2\pi /k= c_1$, which means that: $(\pi/k)^2 =1$, i.e $k=\pm \pi$.

Thanks I solved it now... :-D