[urgent] Eigenfunction again!

[samleemc]

Find the eigenvalue and eigenfunction of the following
\intcos(2X)+cox(2t)y(t) dt = ky , k=eigenvalue , intergrate from 0 to Pi


thx, urgent

this is the original question http://www1.picturepush.com/photo/a/3539509/640/fgcg/a.jpg

[elibj123]

Your equation makes no sense


\int_{0}^{\pi} cos(2t)y(t)dt

Is a number, yet you compare it to a function ( y(t)).

Also

\int_{0}^{\pi} cos(2x)y(t)dt

is an expression of x, and again you compare it to an expression of t ( ky(t))

Are you sure you wrote it correctly?

[samleemc]

Your equation makes no sense


\int_{0}^{\pi} cos(2t)y(t)dt

Is a number, yet you compare it to a function ( y(t)).

Also

\int_{0}^{\pi} cos(2x)y(t)dt

is an expression of x, and again you compare it to an expression of t ( ky(t))

Are you sure you wrote it correctly?

I have uploaded the original version of question, please check!!

[samleemc]

Can anyone please help !?

[HallsofIvy]

What is found in your attachment is
\int_0^\pi (cos(2x)+ cos(2t))y(t)dt= ky(x)
not quite what you originally wrote.

We can write that as
cos(2x)\int_0^\pi y(t)dt+ \int_0^\pi cos(2t)y(t)dt= ky(x)

As elibj123 pointed out, both of those integrals are NUMBERS that, of course, depend on y(x).

Let Y_1= \int_0^\pi y(t)dt
and Y_2= \int_0^\pi cos(2t)y(t).

Now the equation reads simply ky(t)= Y_1 cos(2x)+ Y_2 and we can solve for y(t) just by finding the two numbers Y_1 and Y_2.

Multiply both sides of ky(x)= Y_1 cos(2x)+ Y_2 by cos(2x) and integrate from 0 to \pi:
k\int_0^\pi cos(2x)y(x)dx= Y_1 \int_0^\pi cos^2(2x) dx+ Y_2\int_0^\pi cos(2x)dx

Of course, \int_0^\pi cos(2x)y(x)dx is the same as \int_0^\pi cos(2t)y(t)dt which we have called Y_2. The other two integrals do not involve y(x) and so can be integrated. Rather than do them for you I am going to write \int_0^\pi cos^2(2x)dx= A and \int_0^\pi cos(2x)= B (although that second one ought to be obvious!).

Now, our equation is kY_2= AY_1+ BY_2 or AY_1+ (B- k)Y_2= 0.

If we simply integrate ky(x)= Y_1 cos(2x)+ Y_2 itself from 0 to \pi we get
k\int_0^\pi y(x)dx= Y_1 \int_0^\pi cos(2x)dx+ Y_2\int_0^\pi dx= BY_1+ \pi Y_2
That is the same as kY_1= BY_1+ \pi Y_2 or (B-k)Y_1+ \pi Y_2.

That is, we can solve for y(x) by solving the pair of numerical equations
(B- k)Y_1+ \pi Y_2= 0 and AY_1+ (B- k)Y_2= 0
where A and B are given by the integrals above.

Of course, like any eigenvalue equation, those are satified by the "trivial" solution Y_1= Y_2= 0. Eigenvalues are values of k for which there exist non-trivial solutions. Non-trivial solutions for homogeneous systems of equations occur when the determinant of the coefficient matrix, here (B- k)^2- A\pi, is 0.