# Solving a differential equation based on integrals [closed]

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!

## closed as off-topic by Eevee Trainer, Aqua, RRL, cmk, Adrian KeisterJul 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 ?

• 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 $y(x)=e^x$ :) – dalta Jul 12 at 10:18
• 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 $c= ?.$ – Fred Jul 12 at 10:22
• Got it… $y(x)=-2e^x$. Thanks so much for your help! – dalta Jul 12 at 10:26