Eigenfunction: Find Eigenvalue & Function

[samleemc]

[urgent] Eigenfunction again!

Find the eigenvalue and eigenfunction of the following
[tex]\int[/tex]cos(2X)+cox(2t)y(t) dt = ky , k=eigenvalue , intergrate from 0 to Pi


thx, urgent

this is the original question

 

[elibj123]

Your equation makes no sense[tex]\int_{0}^{\pi} cos(2t)y(t)dt[/tex]

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

Also

[tex]\int_{0}^{\pi} cos(2x)y(t)dt[/tex]

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


[tex]\int_{0}^{\pi} cos(2t)y(t)dt[/tex]

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

Also

[tex]\int_{0}^{\pi} cos(2x)y(t)dt[/tex]

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
[tex]\int_0^\pi (cos(2x)+ cos(2t))y(t)dt= ky(x)[/tex]
not quite what you originally wrote.

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

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

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

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

Multiply both sides of [itex]ky(x)= Y_1 cos(2x)+ Y_2[/itex] by cos(2x) and integrate from 0 to [itex]\pi[/itex]:
[tex]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[/tex]

Of course, [itex]\int_0^\pi cos(2x)y(x)dx[/itex] is the same as [itex]\int_0^\pi cos(2t)y(t)dt[/itex] which we have called [itex]Y_2[/itex]. 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 [itex]\int_0^\pi cos^2(2x)dx= A[/itex] and [itex]\int_0^\pi cos(2x)= B[/itex] (although that second one ought to be obvious!).

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

If we simply integrate [itex]ky(x)= Y_1 cos(2x)+ Y_2[/itex] itself from 0 to [itex]\pi[/itex] we get
[tex]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[/tex]
That is the same as [itex]kY_1= BY_1+ \pi Y_2[/itex] or [itex](B-k)Y_1+ \pi Y_2[/itex].

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

Of course, like any eigenvalue equation, those are satified by the "trivial" solution [itex]Y_1= Y_2= 0[/itex]. 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 [itex](B- k)^2- A\pi[/itex], is 0.