Solve $y'-\int_0^xy(t)dt=2$

Question

I have not idea how to approach this differential equation.

$$y'-\int_0^xy(t)dt=2$$.

Basically, I did,

$$F''(t)-F(x)+F(0)=2 \;\;\;\;\;\;\; F'=y$$

I am stuck.

Thank You.

Answer 1

We differentiate the equation and we find $y''-y=0$, so $$y(x)=Ae^x+Be^{-x}, A,B\in\mathbb{R}$$ and at $x=0$ we have $y'(0)=2$ so we find that $A=B+2$, hence we find $$y(x)=B(e^x+e^{-x})+2e^x,B\in\mathbb{R}.$$

Answer 2

Differentiate, using the Fundamental Theorem of Calculus. You get the probably familiar DE $y'' -y=0$.

Write down the general solution of this DE, and find the values of the constants. Setting $x=0$ will be helpful.

Answer 3

Asuming $y$ is differentiable, you can look at the derived equation $$y''-y=0$$