If $y(t)=1+\int_0^t e^{-(t+v)} y(v)\,dv$ the the value of $y(t)$ at the point $t=1$ is
(A) $0$
(B) $1$
(C) $2$
(D) $3$
I am unable to solve the integral equation using known methods, resolvent kernel, using differentiation.
Using differentiation we get $\frac{\,dy}{\,dt}=1+y+e^{-2t}y$. I'm unable to solve this.
Then how I can I evaluate the value of $y(1)$ ?
If you differentiate (you have a sign error) and solve the initial value problem ($y(0)=1$) numerically, you'll that $y(1)$ has none of the proposed values. Did you post the correct problem? If the question was about the value of $y$ at $t=0$...