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I was given the task to find all continuous functions that satisfy the following equation:

$$x \int_0^x {y }dx=(x+1) \int_0^x{xy}dx$$

I am quite new to differential equations so my first thought was to convert this to a differential equation and then meddle with type of it and how to solve it. So I did the following transformation and then tried to differentiate:

$$\int_0^x {y }dx=(1+\frac{1}{x}) \int_0^x{xy}dx$$ $$y=xy + \frac{1}{x}xy-\frac{1}{x^2} \int_0^x{xy}dx$$

This is where is started having doubts about the correctness of my method. To be honest, I'm not very good with the idea of how to differentiate definite integrals or what does that even mean? What does the expression $\int_0^x{xy}dx$ even mean?

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  • $\begingroup$ what's the domain of $y$, $(0,\infty)$ or $\mathbb{R}$? $\endgroup$ – John May 3 '15 at 16:04
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Since $y=y(x)$, hence I will interpret $\int_0^x{xy}dx$ as $\int_0^x{ty(t)}dt$

Now consider $x \int_0^x {y }dx=(x+1) \int_0^x{xy}dx$. By fundamental theorem of calculus, since $y$ is continuous, you can differentiate both sides w.r.t $x$, which gives you $$ \int_0^x {y}dx+xy=(x+1)xy+ \int_0^x{xy}dx,$$ hence $x^2y(x)=\int_0^x{y}dx-\int_0^x{xy}dx$

Note if $x\not=0$, by fundamental theorem of calculus, $y$ is differentiable. Hence differentiate both sides again, $2xy+x^2y'=y-xy$. This you can solve $y$ by standard ODE method.

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  • $\begingroup$ Thanks :D But what happened to $x$ in front of $\int_0^x{ydx}$ ?? $\endgroup$ – Transcendental May 3 '15 at 16:08

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