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 '16 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 '16 at 5:58
  • $\begingroup$ There is some error in my differential equation. I shall correct and revert back. $\endgroup$ – Tosh Sep 22 '16 at 5:59
  • $\begingroup$ I got it. I shall post as an answer $\endgroup$ – Tosh Sep 22 '16 at 6:05

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

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 '16 at 6:32

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