I have the Fredholm integro-differential equation $$u'(x)= u(x) - \frac{x}{2} + \frac{1}{x+1} - \ln(x+1) + \frac{1}{\ln^2 2} \int_0^1 \frac{x}{t+1} u(t)dt, \quad u(0)=0.$$ and I know the exact solution $$u(x)=\ln(x+1).$$
I need to convert this equation to a second order IVP. The related question is here.
My attempt is below:
Differentiating in $x$ the both sides of equation and the extract solution: $$u''(x) - u'(x)= -\frac{1}{2} - \frac{1}{(x+1)^2}-\frac{1}{x+1} + \frac{1}{\ln^2 2} \int_0^1 \frac{u(t)}{t+1}dt, \quad u(0)=0, \quad u'(0)=1.$$
Question. How to write the second IVP without the integral? Should use I the partial integration?
First by rearrangement of terms we have:$$u'-\frac{1}{x+1}=(u-\ln{(x+1)})-\frac{x}{2}+\frac{x}{\ln^2 2}\int_{0}^{1}\frac{u(t)}{t+1}dt$$ by defining $w=u-\ln(x+1)$ we get:$$w'=w-\frac{x}{2}+\frac{x}{\ln^2 2}\int_{0}^{1}\frac{w(t)+\ln(t+1)}{t+1}dt=w-\frac{x}{2}+\frac{x}{\ln^2 2}\int_{0}^{1}\frac{w(t)}{t+1}dt+\frac{x}{\ln^2 2}\int_{0}^{1}\frac{\ln(t+1)}{t+1}dt$$ where $\int_{0}^{1}\frac{\ln(t+1)}{t+1}dt=\frac{\ln^2 2}{2}$. Then by substitution we finally arrive at:$$w'(x)=w(x)+\frac{x}{\ln^2 2}\int_{0}^{1}\frac{w(t)}{t+1}dt$$Now differentiating leads us to:$$w''-w'=\int_{0}^{1}\frac{w(t)}{t+1}dt$$ also we had $$\int_{0}^{1}\frac{w(t)}{t+1}dt=\frac{w'-w}{x}$$then by substitution we obtain:$$x(w''-w')-(w'-w)=0$$Let $v=w'-w$ therefore $$xv'-v=0$$ which leads us to $v=w'-w=cx$ for constant $c$ and again solving this ODE gives us $w=c(e^x-x-1)$ and finally: $$\large u(x)=\ln(x+1)+c(e^x-x-1)$$