$$\varphi(x) - \dfrac {\pi^2} {4}\int\limits_0^1 K(x,t)\varphi(t) \: dt=\dfrac{x}{2}$$ $$K(x,t) = \begin{array}{c} \dfrac{x(2-t)} {2} \quad 0\leq x <t\\ % just use \\ \dfrac{t(2-x)} {2} \quad t\leq x <1 \end{array} $$
I am helping my friend solving this integral equation but I took this lesson way back and I dont remember much of it. What I have done here is $$\varphi(x) = \dfrac{x} {2}+\dfrac {\pi^2} {8}\cdot(2-x) \int\limits_0^x t\cdot\varphi(t) \: dt +\dfrac {\pi^2} {8}\cdot t \int\limits_x^1 (2-t)\cdot\varphi(t) \: dt $$
$$\varphi'(x) = \dfrac{1} {2}-\dfrac {\pi^2} {8} \int\limits_0^x t\cdot\varphi(t) \: dt +\dfrac {x\pi^2} {8}\varphi(t)+\dfrac {\pi^2} {8} \int\limits_x^1 (2-t)\cdot\varphi(t) \: dt $$ $$\varphi''(x) = \dfrac{x\pi^2}{8}\varphi'(x)$$
what I can do next ?
or is there any other way to solve this equation ?
Continuing form where I left off in my comment, we have the ODE $$ \psi' = \frac{x\pi^2}{8}\psi, $$ which has general solution $\psi(x) = \alpha e^{\pi^2x^2/16}$. Therefore $$\varphi(x) = \frac{\alpha\pi^{3/2}}{8}\mathrm{erfi}\left(\frac{4}{\pi}x\right)+\beta,$$ where $\mathrm{erfi}$ is the imaginry error function and $\alpha$ and $\beta$ are two undetermined constants.
At $x=0$, the integral equation reads $\phi(0) = 0$ since the kernel is $0$ when $x=0$. This implies that $\beta = 0$. $\alpha$ seems much more difficult to obtain since it requires you to evaluate the integral involving the kernel and $\mathrm{erfi}$. Setting $x=1$ is probably the most natural choice, as then you will have something like $t\cdot \mathrm{erfi}(stuff*t)$ and you maybe be able to may be able to simplify using integration by parts but I doubt you will be able to get a closed form.