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Given

\[

f(x) = 1 + \lambda\int_0^1(xy + x^3y^2)f(y)dy.

\]

To use the Fredholm series, we need to find \(D(\lambda)\) wich is \(1 - \text{order }\lambda + \cdots\). I have already calculated the orders and of order 3 and 4 the terms are 0 so the terms greater than 3 are all zero since we have repeated zeros.

\[

\text{order }\lambda:\quad\int_0^1K(y,y)dy = \frac{1}{2}

\]

and

\[

\text{order }\lambda^2:\quad\iint_0^1

\begin{vmatrix}

K(y,y) & K(y,y')\\

K(y',y) & K(y',y')

\end{vmatrix}

dydy' = \frac{1}{90}

\]

Therefore,

\[

D(\lambda) = 1 - \frac{\lambda}{2} + \frac{\lambda^2}{90}.

\]

Next we need to find \(\mathcal{D}(x,y;\lambda)\). In the Taylor series, we obtain repeated zeros after order lambda squared and lambda cubed. Again, repeated zeros so

\[

\mathcal{D}(x,y;\lambda) = K(x,y) - \lambda\int_0^1

\begin{vmatrix}

K(x,y) & K(x,z)\\

K(z,y) & K(z,z)

\end{vmatrix}dz =

-\lambda \left(\frac{x^3 y^2}{3}-\frac{x^3 y}{4}-\frac{x y^2}{5}+\frac{x y}{6}\right)+x^3 y^2+x y

\]

I get

\[

f(x) = 1 + \frac{\lambda}{D(\lambda)}\int_0^1\mathcal{D}(x,y;\lambda)dy

= \text{see second post}

\]

\[

f(x) = 1 + \lambda\int_0^1(xy + x^3y^2)f(y)dy.

\]

To use the Fredholm series, we need to find \(D(\lambda)\) wich is \(1 - \text{order }\lambda + \cdots\). I have already calculated the orders and of order 3 and 4 the terms are 0 so the terms greater than 3 are all zero since we have repeated zeros.

\[

\text{order }\lambda:\quad\int_0^1K(y,y)dy = \frac{1}{2}

\]

and

\[

\text{order }\lambda^2:\quad\iint_0^1

\begin{vmatrix}

K(y,y) & K(y,y')\\

K(y',y) & K(y',y')

\end{vmatrix}

dydy' = \frac{1}{90}

\]

Therefore,

\[

D(\lambda) = 1 - \frac{\lambda}{2} + \frac{\lambda^2}{90}.

\]

Next we need to find \(\mathcal{D}(x,y;\lambda)\). In the Taylor series, we obtain repeated zeros after order lambda squared and lambda cubed. Again, repeated zeros so

\[

\mathcal{D}(x,y;\lambda) = K(x,y) - \lambda\int_0^1

\begin{vmatrix}

K(x,y) & K(x,z)\\

K(z,y) & K(z,z)

\end{vmatrix}dz =

-\lambda \left(\frac{x^3 y^2}{3}-\frac{x^3 y}{4}-\frac{x y^2}{5}+\frac{x y}{6}\right)+x^3 y^2+x y

\]

I get

\[

f(x) = 1 + \frac{\lambda}{D(\lambda)}\int_0^1\mathcal{D}(x,y;\lambda)dy

= \text{see second post}

\]

Last edited: