Solve the integral equation $\displaystyle \int_0^x(1-x^2+t^2)\phi(t)\,dt=\frac{x^2}{2}$.
On differentiation, we get $$\phi(x)=x+2\int_0^x x\phi(t)\,dt.$$
I'm unable to find resolvent kernal so that I can solve. Also not able to put it in convolution form. How can I proceed further?
On
Hint for a start
Starting with $$\displaystyle \int_0^x(1-x^2+t^2)\,\phi(t)\,dt=\frac{x^2}{2}$$ differentiate a first time with respect of $x$ to get $$-2x\int_0^x \phi (t) \, dt+\phi (x)=x$$ Do it again $$-2\int_0^x \phi (t) \, dt+\phi '(x)-2 x \phi (x)=1$$ and again $$\phi ''(x)-2 x \phi '(x)-4 \phi (x)=0$$
On
HINT-As Claude Leibovici suggested to get $$\phi ''(x)-2 x \phi '(x)-4 \phi (x)=0$$
Multiply by $x$ to observe $x\phi''(x)=d(2x^2\phi(x))$
Integrate to get $2x^2\phi(x)=x\phi'(x)-\phi(x)$
or $(2x^2+1)\phi(x)=x\phi'(x)\implies\frac{\phi'(x)}{\phi(x)}=(2x+1/x)\implies In(\phi(x))=x^2+In(x)+a, etc$
For $x\neq 0$ rewrite \begin{equation*} \phi(x)=x+2\int_{0}^{x}x\phi(t)\,dt \end{equation*} as \begin{equation*} \psi(x)=1+2\int_{0}^{x}t\psi(t)\,dt\tag{1} \end{equation*} where $\psi(x) = \dfrac{\phi(x)}{x}$ and $\lim_{x\to 0}\psi(x) = 1$. Differentiation of $(1)$ yields \begin{equation*} \psi'(x)=2x\psi(x) \Longleftrightarrow \psi(x) = Ce^{x^2}. \end{equation*} According to the limit $C=1.$
Consequently \begin{equation*} \phi(x) = xe^{x^2} \end{equation*} which satisfies the original integral equation.