MHB Integral equation Fredholm series

[Dustinsfl]

Given
\[
f(x) = 1 + \lambda\int_0^1(xy + x^3y^2)f(y)dy.
\]
To use the Fredholm series, we need to find \(D(\lambda)\) which 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
\[
f(x) = 1 + \frac{\lambda}{D(\lambda)}\int_0^1\mathcal{D}(x,y;\lambda)dy
= \text{see second post}
\]

 

[Dustinsfl]

The solution I get for the Fredholm method is
\[
f(x) = 1 - 3x\lambda\frac{\lambda - 30}{180 + \lambda(2\lambda - 45)} +
5x^3\lambda\frac{\lambda + 24}{360 + 2\lambda(2\lambda - 45)},
\]
and when I use the sum of separable kernels, I get
\[
f(x) = 1 - 3x\lambda\frac{\lambda - 30}{180 + \lambda(\lambda - 90)} +
5x^3\lambda\frac{\lambda + 24}{360 + 2\lambda(\lambda - 90)},
\]
The math for the Fredholm method is above and the math for the separable kernels is in http://mathhelpboards.com/differential-equations-17/integral-equations-7278.html second post. They should yield the same result but aren't. They are just off a little. Which one is wrong? Where is there a mistake?