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I need to solve the following ODE

$\mu p_tF^\prime(p_t)+\frac{1}{2}\sigma^2p_t^2F^{\prime\prime}(p_t)-rF(p_t)+Ap_t+b=0$

I know a general solution is on the form



$\beta_1=\frac{-(\mu-\frac{1}{2} \sigma^2)+\sqrt{(\mu-\frac{1}{2}\sigma^2)^2+2\sigma^2r}}{\sigma^2}$

$\beta_2=\frac{-(\mu-\frac{1}{2} \sigma^2)-\sqrt{(\mu-\frac{1}{2}\sigma^2)^2+2\sigma^2r}}{\sigma^2}$

and in the article I am told that $\beta$'s are a solution to the characteristic equation

$\frac{1}{2}\sigma^2p(p-1)+\mu p -r=0$

But I cannot seem to figure out how I get this characteristic equation?

Any help is deeply appreciated

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up vote 1 down vote accepted

This is a Cauchy–Euler equation: $$x^2y''+\frac{2\mu}{\sigma^2}xy'-\frac{2r}{\sigma^2}y=2\frac{-Ax+b}{\sigma^2}$$ (I denoted $y=F(p_t)$, $x=p_t$ and multiplied by $\frac{2}{\sigma^2}$) The characteristic equation comes from the substitution $y=x^p$ to the homogeneous equation $$x^2y''+\frac{2\mu}{\sigma^2}xy'-\frac{2r}{\sigma^2}y=0$$ We have $$\begin{align*}0&=x^2 p(p-1)x^{p-2}+\frac{2\mu}{\sigma^2}xpx^{p-1}-\frac{2r}{\sigma^2}x^p\\&=x^p\left[p(p-1)+\frac{2\mu}{\sigma^2}p-\frac{2r}{\sigma^2}\right]=\frac{2x^p}{\sigma^2}\left[\frac{\sigma^2}{2}p(p-1)+\mu p-r\right]\end{align*}$$ So $p$ is a solution to $\frac{\sigma^2}{2}p(p-1)+\mu p-r=0$

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If you have

$a t^2 F^{(2)}(t) + b t F^{(1)}(t) + c F(t) = 0$

Then suppose $F(t) = t^x$ is the form of a solution Then $F^{(1)}(t) = x t^{x-1} = x \frac{f}{t}$ and $F^{(2)}(t) = x(x-1) t^{x-2} = x(x-1) \frac{f}{t^2}$

Substituting back into the first equation gets:

$a x(x-1) f + b x f + c f = 0$ So $a x(x-1) + b x + c = 0$

The characteristic equation is solving for x

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