If $y$ is a function of $x$ then solve
The integro-differential equation $$x^2 \frac{dy}{dx}+\int_{0}^{x} t^2y^5(t)dt=0$$
I tried to differentiate both sides w.r.t $x$ we get
$$x^2 \frac{d^2y}{dx^2}+2x\frac{dy}{dx}+x^2y^5=0$$ $\implies$
$$x\frac{d^2y}{dx^2}+2\frac{dy}{dx}+xy^5=0$$
Letting $p=\frac{dy}{dx}$ we have a Linear differential equation as:
$$\frac{dp}{dx}+\frac{2p}{x}=-y^5$$ using Integrating factor method we get
$$px^2=-\int x^2y^5dx+C$$
How to proceed from here?
In Polyanin A., D., Zaitsev V. F., Handbook of exact solutions for ordinary differential equations the ODE \begin{equation} x\frac{d^2y}{dx^2}+2\frac{dy}{dx}+xy^5=0 \end{equation} is mentioned as a solvable case of a modified Emden-Fowler Equation: \begin{equation} xy''+\frac{2n+m+3}{m-1}y'=Ax^{n+1}y^m \end{equation} Here, $n=0,m=5,A=-1$. Its solution in parametric form is (2.6.3): \begin{align} x&=\exp\left( -2C_2\int\frac{dt}{\sqrt{C_1+t^2/4-t^6/3}} \right)\\ y&=t\exp\left( C_2\int\frac{dt}{\sqrt{C_1+t^2/4-t^6/3}} \right) \end{align} It can be checked by introducing the form $x=\exp(g(t)),y=t\exp(-g(t)/2)$ in the initial ODE. Then $g(t)$ verifies $$ 4g''(t)-t(4t^4-1)[g'(t)]^3=0$$