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$
Dividing by $(-x)$
$ \frac{dv}{dx}+axv=-\frac{b}{x}v$
$ \frac{dv}{dx}+\frac{b}{x}v=-axv$
I.F
$e^{\int \frac{b}{x} dx}=x^b$
$\therefore vx^b=b\int x^ba dx$
$\therefore v= \frac{a}{b+1}x+Cx^{-b}$
$U=\int \frac{a}{b+1}x+Cx^{-b}$
But the answer is
$U(x)=c\int^x_0 t^{-b}e^{-at}dt$
I cannot understand where the $t$ comes to sit in the equation.
$$-x \frac{d^2U}{dx^2}=(ax+b)\frac{dU}{dx}$$
Letting $\frac{dU}{dx}=v$
$$-x \frac{dv}{dx}=(ax+b)v$$
Rearranging
$$-\frac{x}{v} \frac{dv}{dx}=(ax+b)$$
$$ -\int\frac{1}{v} dv= \int (a+\frac{b}{x}) dx $$
$$ v=e^{-ax}x^b$$
By taking $t$ as a dummy variable.
$$U(x)= \int ^x_0(e^{-at}t^{-b}) dt$$