Solve the integeral equation (C.S.I.R) Let $\lambda_1, \lambda_2$ be the eigen value and $f_1 , f_2$ be the coressponding eigen  functions for the homogeneous integeral equation
$$ \phi(x) - \lambda \int_0^1 (xt +2x^2) \phi(t) dt = 0 $$
Then


*

*$\lambda_1 = -18 - 6 \sqrt{10} , \lambda_2 = -18 + 6 \sqrt{10}$

*$\lambda_1 = -36 - 12 \sqrt{10} , \lambda_2 = -36 + 12 \sqrt{10}$

*$\int_0^1f_1(x)f_2(x) dx = 1$

*$\int_0^1f_1(x)f_2(x) dx = 0$
I have solved simply and get the values of $\lambda_1 \ \ and \ \ \lambda_2$ belong to (1).
Please tell me about (3) and (4) option.
Thank you.
 A: Let's rewrite your equation in the following form:
\begin{align}
\frac{1}{\lambda}\phi(x) = x\int_{0}^{1}t\phi(t) \mathrm{d}t + 2x^{2}\int_{0}^{1} \phi(t) \mathrm{d}t = c_{1}x + c_{2}x^{2}
\end{align}
for $c_{1}: = \int_{0}^{1}t\phi(t) \, \mathrm{d}t$ and $c_{2}:= 2\int_{0}\phi(t)\, \mathrm{d}t$. From this we can see that $\phi$ is a quadratic polynomial with no constant term.
Substituting $\phi(x) = c_{1}x + c_{2}x^{2}$, we find that 
\begin{align}
\frac{c_{1}}{\lambda} x + \frac{c_{2}}{\lambda}x^{2} 
&=x \int_{0}^{1} t(c_{1}t+ c_{2}t^{2}) \, \mathrm{d}t + 2x^{2} \int_{0}^{1} c_{1}t + c_{2}t^{2} \, \mathrm{d}t \\
&=x(c_{1}/3 + c_{2}/4) + 2x^{2}(c_{1}/2 + c_{2}/3)
\end{align}
so that 
\begin{align}
\frac{c_{1}}{\lambda} &= \frac{c_{1}}{3} + \frac{c_{2}}{4} \\
\frac{c_{2}}{\lambda} &= c_{1} + \frac{2c_{2}}{3} \\
\end{align}
It can be verified that if both $c_{1}$ and $c_{2}$ are nonzero, then $\lambda = -18 \pm 6\sqrt{10}$. 
Since $c_{1} = \left(\frac{1}{\lambda} - \frac{2}{3}  \right)c_{2}$
\begin{align}
f_{1}(x) &= -\left(\frac{1}{18+6\sqrt{10}} + \frac{2}{3}\right)c_{2}x + c_{2}x^{2}
= -\frac{1}{6}(1+\sqrt{10})c_{2}x + c_{2}x^{2} \\
f_{2}(x) &= -\left(\frac{1}{18-6\sqrt{10}} + \frac{2}{3}\right)c_{2}'x + c_{2}'x^{2} = -\frac{1}{6}(1-\sqrt{10})c_{2}'x + c_{2}'x^{2}
\end{align}
so that 
\begin{align}
\int_{0}^{1} f_{1}(x)f_{2}(x) \, \mathrm{d}x
&=
c_{2}c_{2}' \int_{0}^{1} x^{4} -\frac{1}{3}x^{3} - \frac{1}{4}x^{4} \, \mathrm{d}x = \frac{c_{2}c_{2}'}{30}
\end{align}
So we can't claim that these eigenfunctions are orthogonal, or that their inner product is 1, unless we choose a scaling such that $c_{2}c_{2}' = 30$. 
A: This work is almost completed.
Define $g(t)=\int_0^t \phi(s) ds$ and $h(t)=\int_0^t s \phi(s) ds$
Because $\phi(0)=0$, we have $g(0)=0$ and $h(0)=0$.
Using integration by parts, we have
$$h(t)=\int_0^t t \phi(t) dt=t\left(\int_0^t\phi(s) ds\right)-\int_0^t\left(\int_0^s \phi(r) dr\right)ds$$
$$\implies h(t)=tg(t)-\int_0^t g(s) ds......(1)$$
The original equation becomes:
$$\phi(x) = \lambda x\ h(1) +2 \lambda x^2 g(1)......(2)$$
And
$$g(t)=\int_0^t \phi(s) ds=\frac{1}{2}\lambda h(1)t^2 +\frac{2}{3} \lambda g(1) t^3......(3)$$
$$h(t)=\int_0^t s\phi(s) ds=\frac{1}{3}\lambda h(1)t^3 +\frac{2}{4} \lambda g(1) t^4......(4)$$
Therefore
$$g(1)=\frac{1}{2} \lambda h(1)+\frac{2}{3} \lambda g(1)...(5)$$
$$h(1)=\frac{1}{3}\lambda h(1) +\frac{2}{4} \lambda g(1)......(6)$$
Setting $\mu=\lambda^{-1}$ in (5) and (6), we have:
$$
        \left(\begin{matrix} \frac{2}{3} & \frac{2}{3}  \\ \frac{2}{4} & \frac{1}{3}  \\ \end{matrix}\right)
        \left(\begin{matrix} g(1)  \\ h(1)   \\ \end{matrix}\right)=
        \mu \left(\begin{matrix} g(1)  \\ h(1)   \\ \end{matrix}\right)......(7)
$$
Once you found two eigen values you can substitute them into (7) and solve for $g(1)$ as a function of $h(1)$.  You may then fix $h(1)$ by some kind of normalization.
Finally you can substitute $\lambda_{1,2}$ and $g(1),h(1)$ into (2) to obtain the eigen functions $f_1(x)$ and $f_2(x)$.
I hope that you can pick it up from here.
-mike
