Solve the integral equation displaystyle∈t_0^x(x-t)y(t)…

Solve the integral equation $\displaystyle\int_0^x(x-t)y(t)\,dt=2x+\int_0^x y(t)\,dt$.

Step 1: Split and differentiate once

Write $\displaystyle\int_0^x(x-t)y(t)\,dt=x\!\int_0^x y\,dt-\!\int_0^x ty\,dt$. With $I(x)=\int_0^x y\,dt$, the equation becomes $(x-1)I(x)-\int_0^x ty\,dt=2x$. Differentiating (using $I'=y$, and $\frac{d}{dx}\int_0^x ty\,dt=xy$):

\[I(x)+(x-1)y-xy=2\quad\Longrightarrow\quad I(x)-y(x)=2.\]

Step 2: Differentiate again

Differentiating $I(x)-y(x)=2$ gives $y-y'=0$, i.e. $y'=y$, so $y=Ae^x$.

Step 3: Determine the constant

Substitute $y=Ae^x$ into $I(x)-y(x)=2$: $\displaystyle\int_0^x Ae^t\,dt-Ae^x=A(e^x-1)-Ae^x=-A=2$, so $A=-2$.