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It is my first week dealing with Differential Equations, and I am totally lost at solving the following equation:

$\int^x_0(x-t)y(t)dt=2x+\int^x_0y(t)dt$

Any help would be greatly appreciated!

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closed as off-topic by Eevee Trainer, Aqua, RRL, cmk, Adrian Keister Jul 12 at 16:59

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From $\int^x_0(x-t)y(t)dt=2x+\int^x_0y(t)dt$ we derive

$$x\int^x_0y(t)dt -\int_0^xty(t)dt=2x+\int^x_0y(t)dt.$$

If we differentiate we get

$$\int^x_0y(t)dt+xy(x)-xy(x)=2+y(x).$$

Hence

$$\int^x_0y(t)dt=2+y(x).$$

Differentiation once again yields

$$y(x)=y'(x).$$

Can you proceed ?

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  • $\begingroup$ Thank you so much, Fred. Your answer explains the methodology perfectly. And yeah, based on my one lecture dealing with the topic, I assume that this result would mean that $y(x)=e^x$ :) $\endgroup$ – dalta Jul 12 at 10:18
  • $\begingroup$ Be careful ! The differential equation $y(x)=y'(x)$ has the general solution $y(x)=ce^x$. From $\int^x_0y(t)dt=2+y(x)$ we deduce that $y(0)=-2.$ Hence $c= ?.$ $\endgroup$ – Fred Jul 12 at 10:22
  • $\begingroup$ Got it… $y(x)=-2e^x$. Thanks so much for your help! $\endgroup$ – dalta Jul 12 at 10:26

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