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I am trying to solve a ODE that arises from a Hamilton-Jacobi-Bellman (HJB) equation. The equation is $$\frac{1}{2}b^2(1-\rho_s^2)\psi''-\frac{1}{2}\left(\frac{\mu-r}{\sigma}\right)^2\frac{(\psi')^2}{\psi''}+[ru+\theta a+b\rho_s(\mu-r)(1-\frac{2}{\sigma})]\psi'=0,$$ where $\mu, r, \sigma, \theta, a, \rho_s, b$ are constant. I want to determine $\psi'(u)$ (so that I get an integral form for $\psi(u)$). I have tried guessing (trial and error method) forms of the solution but didn't get far. I also tried the Legendre transform, but could not get the linear form. These are the method that I have seen being used in with these problems.

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Apart from $u$ and $\psi$, the other variables are constant? –  Aryabhata Sep 15 '10 at 21:02
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1 Answer 1

up vote 2 down vote accepted

Assume $\psi' = f$, then the equation is of the form:

$$Af' + B\frac{f^2}{f'} + (cu+d)f = 0$$

By putting $$g = \frac{f'}{f} = (\log f)'$$ we see that

$$Ag + \frac{B}{g} + (cu+d) = 0$$

This is a quadratic in $g$ and can be easily solved.

We get $$ \psi' = f = e^{\int g}$$

Hope that helps.

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thanx, you make it to be so simple. –  Vaolter Sep 16 '10 at 11:07
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