Eigenvalue and eigenfunction for Fredholm method

[Dustinsfl]

Given
\[
f(x) = \lambda\int_0^1xy^2f(y)dy
\]
At order \(\lambda^2\) and \(\lambda^3\), we have repeated zeros so
\[
D(\lambda) = 1 - \frac{\lambda}{4}.
\]
Then we have
\[
\mathcal{D}(x, y;\lambda) = xy^2
\]
so
\[
f(x) = \frac{\lambda}{D(\lambda)}\int_0^1\mathcal{D}(x, y;\lambda)dy.
\]
How do I get the eigenfunction and value from this method?

 

[gtoguy97]

The eigenfunction and eigenvalue can be found by solving the following equation:\[f(x) = \lambda f(x),\]where \(\lambda\) is the eigenvalue and \(f(x)\) is the eigenfunction. Substituting the expression for \(f(x)\) into the equation, we get \[\frac{\lambda}{D(\lambda)}\int_0^1\mathcal{D}(x, y;\lambda)dy = \lambda f(x).\]We can rearrange this equation to solve for \(\lambda\):\[\lambda = \frac{D(\lambda)\int_0^1\mathcal{D}(x, y;\lambda)dy}{f(x)}.\]Once we have solved for \(\lambda\), we can substitute it back into the expression for \(f(x)\) to find the eigenfunction:\[f(x) = \frac{\lambda}{D(\lambda)}\int_0^1\mathcal{D}(x, y;\lambda)dy.\]