I have the Fredholm integro-differential equation $$u'(x)= u(x) + \int_0^1 \frac{x}{t+1} u(t)dt, \quad u(0)=0.$$
I need to find the corresponded second order IVP.
My attemp is:
Differentiating in x the both sides of equation and inital condition: $$u''(x) - u'(x)= \int_0^1 \frac{u(t)}{t+1}dt, \quad u(0)=0, \quad u'(0)=0.$$
Question. How to be with the integral $\int_0^1 \frac{u(t)}{t+1}dt$ here? Is the $u'(0)$ equal to $0$?
What you have arrived at is true. Now by substituting we get:$$u''-u'=\int_0^1 \frac{u}{t+1}dt=\frac{u'-u}{x}$$ which by replacing $w=u'-u$ leads to $$xw'-w=0$$It's easy to see that the answer is $w=ct$ for constant $c$. Then we have$$u'-u=cx$$. The answer has been divided to two parts: homogeneous and private where private one is $-ct-c$ and the homogeneous one is $ae^t$. By replacing and imposing initial conditions we get $a=c$ and $u=c(e^t-t-1)$