Integral form of this IVP

Question

How do I show that the following initial value problem $$ xu''+u'+xu=0,\quad u(0)=1,\quad u'(0)=0 $$ has the following integral form: $$ u(x)=1+\int_{0}^{x} t\ln(t/x)u(t)\,dt $$ I am stuck because if I divide both sides of both ODE by $x$ $$ u'+\frac{1}{x}u+u=0 $$ $\frac{1}{x}$ is undefined at $0$.

Answer 1

Hint: From $$u(x)=1+\int_{0}^{x} t\ln(t/x)u(t)\,dt,$$ calculate $u'(x)$ and $u''(x)$ and show that this satisfies $xu''+u'+xu=0$. Also show that $u(0) = 1$ and $u'(0) = 0$ (using L'Hopital's rule, if necessary).