So I just got out of my first exam on differential equations. One point had an integral equation which we were supposed to turn into a first order linear differential equation. The thing is I wasn't able to get rid of the integral and ended with an integro differential equation that looks like this:
$\frac{df}{dx}=\frac{1}{nx}f-\frac{1}{nx^2}\int_{0}^{x}f(k)dk ;~ n\in \mathbb{R}$.
I'm not sure if I made any mistakes, but that's not the point. How could I solve an equation like that?
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Turn the equation into $$nx^2\frac {df}{dx}=xf-\int_0^xf(k)dk$$
and differentiate (using Fundamental Theorem of Calculus):
$$nx^2\frac {d^2f}{dx^2}+2nx\frac{df}{dx}=f+x\frac {df}{dx}-f=x\frac{df}{dx}$$
Then simplify, and let $g(x)=\dfrac {df}{dx}$:
$$\frac {dg}{dx}=\frac {1-2n}{nx}g$$
which is a first-order ODE, with solution:
$$g(x)=c_1x^{\frac1n-2}\implies f(x)=\begin{cases}\frac {c_1n}{1-n}x^{\frac1n-1}+c_2&\text{for }n\ne 1\\c_1\ln x + c_2&\text{for }n=1\end{cases}$$
$$\dfrac{df}{dx}=\dfrac{f}{nx}-\frac{1}{nx^2}\int_{0}^{x}f(k)dk ;~ n\in \mathbb{R}$$ Note that: $$nx^2\frac {df}{dx}=xf-\int_0^xf(k)dk$$ $$n\frac {df}{dx}=\left ( \dfrac {\int_0^xf(k)dk}x \right )'$$ Integrate both sides.