Find the eigenvalues and eigenfunction for the BVP:
y ' ' '+ λ2 y ' = 0 , y(0) = 0, y’(0) = 0 and y’(L) = 0
To find the eigenvalues and eigenfunctions for a given boundary value problem (BVP), we need to solve the differential equation and apply the boundary conditions.
The given differential equation is:
y'' + λ^2y' = 0
Let's solve this equation by assuming a solution of the form y(x) = e^(rx), where r is a constant.
Taking the derivatives, we have:
y' = r e^(rx)
y'' = r^2 e^(rx)
Substituting these into the differential equation, we get:
r^2 e^(rx) + λ^2 r e^(rx) = 0
Factoring out e^(rx), we have:
e^(rx)(r^2 + λ^2 r) = 0
Since e^(rx) is never zero, we can divide both sides by e^(rx):
r^2 + λ^2 r = 0
This is a quadratic equation in r. To solve it, we can use the quadratic formula:
r = (-λ^2 ± sqrt(λ^4 - 4*1*0)) / 2*1
Simplifying, we have:
r = (-λ^2 ± sqrt(λ^4)) / 2
r = (-λ^2 ± λ^2) / 2
There are two roots for r:
r1 = 0
r2 = -λ^2
Now, we can find the eigenvalues by setting the expression for r2 = -λ^2 equal to zero and solving for λ:
-λ^2 = 0
λ = 0
Therefore, the eigenvalues, λ, are 0 and -λ^2.
To find the corresponding eigenfunctions, we substitute the values of r1 and r2 into y(x) = e^(rx).
For r1 = 0, y(x) = e^(0*x) = e^0 = 1
So, the eigenfunction corresponding to λ = 0 is y(x) = 1.
For r2 = -λ^2 = -(-λ^2) = λ^2, y(x) = e^(λ^2*x)
So, the eigenfunction corresponding to λ = -λ^2 is y(x) = e^(λ^2*x).
Therefore, the eigenfunctions for the given BVP are y(x) = 1 and y(x) = e^(λ^2*x), where λ can be any real number.