A utility function $U$ whose corresponding relative risk aversion function is a linear, increasing function satisfies the differential equation


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 $\frac{dU}{dx}=v$

$$-x \frac{dv}{dx}=(ax+b)v$$

$$x \frac{dv}{dx}+axv=-bv$$


$$e^{\int (ax) dx}=e^{a\frac{x^2}{2}}$$

$$\therefore ve^{a\frac{x^2}{2}}=b\int e^{a\frac{x^2}{2}} dx$$

$$\therefore ve^{a\frac{x^2}{2}}=ba^2x e^{a\frac{x^2}{2}}+C$$



$$u=ba^2\frac{x^2}{2}+C\int e^{-a\frac{x^2}{2}} dx$$

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.

  • $\begingroup$ Are u sure $a^{-at}$ ? What is your reference? $\endgroup$
    – user16651
    Sep 22, 2016 at 5:52
  • $\begingroup$ @BehrouzMalekiit is $e^{-at}$. It is a homework my teacher gave. I don't know from where she got it. $\endgroup$
    – Tosh
    Sep 22, 2016 at 5:58
  • $\begingroup$ There is some error in my differential equation. I shall correct and revert back. $\endgroup$
    – Tosh
    Sep 22, 2016 at 5:59
  • $\begingroup$ I got it. I shall post as an answer $\endgroup$
    – Tosh
    Sep 22, 2016 at 6:05

1 Answer 1


$$-x \frac{d^2U}{dx^2}=(ax+b)\frac{dU}{dx}$$

Letting $\frac{dU}{dx}=v$

$$-x \frac{dv}{dx}=(ax+b)v$$


$$-\frac{1}{v} dv=\left(a+\frac{b}{x}\right)dx$$

$$ -\int(\ln{v}) dv= \int \left(a+\frac{b}{x}\right) dx $$

$$ v=C e^{-ax}x^b$$

Where $C$ is a constant. By taking $t$ as a dummy variable.

$$U(x)= C\int ^x_0e^{-at}t^{-b}\, dt$$

  • 1
    $\begingroup$ Indeed your question wasn't about quantitative finance. $\endgroup$
    – user16651
    Sep 22, 2016 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.