A utility function $U$ whose corresponding relative risk aversion function is a linear, increasing function satisfies the differential equation
$$-x\frac{U''(x)}{U'(x)}=ax+b$$
for some constants $a>o$ and $b\in \mathbf{R}$
Show that
$$U(x)=c \int _0 ^x t^{-b}e^{-at} dt$$ where $c>0$ is an arbitrary constant.
I massaged the equation so it becomes friendly.
$$-x \frac{d^2U}{dx^2}=(ax+b)\frac{dU}{dx}$$
Letting $\frac{dU}{dx}=v$
$$-x \frac{dv}{dx}=(ax+b)v$$
$$x \frac{dv}{dx}+axv=-bv$$
I.F
$$e^{\int (ax) dx}=e^{a\frac{x^2}{2}}$$
$$\therefore ve^{a\frac{x^2}{2}}=b\int e^{a\frac{x^2}{2}} dx$$
$$\therefore ve^{a\frac{x^2}{2}}=ba^2x e^{a\frac{x^2}{2}}+C$$
$$v=ba^2x+Ce^{-a\frac{x^2}{2}}$$
$$\frac{dU}{dx}=ba^2x+Ce^{-a\frac{x^2}{2}}$$
$$u=ba^2\frac{x^2}{2}+C\int e^{-a\frac{x^2}{2}} dx$$
But the answer is
$$U(x)=c\int^c_0 t^{-b}e^{-at}dt$$
I cannot understand where the $t$ comes to sit in the equation.