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$rV = \pi x − f + \mu x \frac{\partial V(x)}{\partial x} + 0.5\sigma^2 x^2 \frac{\partial^2 V(x)}{\partial x^2}$

Why is the general solution given by: $V(x) = A_0 + A_1 x + A_2 x^\lambda + A_3 x^\beta$?

I thought that since this is a second order non-homogenous differential equation, that the solution would consist of a particular + general solution, where the general solution would look something like $V(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ where $r_i$ the roots of the corresponding polynomial.

Could anyone please explain? Thanks in advance.

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2 Answers 2

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This is called a Cauchy–Euler equation. The important point is that the coefficients are not constant, but the terms containing derivatives are all of the form $$ x^k \frac{d^k}{dx^k} V, $$ which means that substituting $y=\log{x}$, $$ \frac{d}{dy} = \frac{1}{x} \frac{d}{dx}, \quad \text{or} \quad \frac{d}{dx} = e^y \frac{d}{dy} $$ which allows you to reduce it to a constant-coefficient equation (but beware that derivatives higher than the first create more than one term when you use the chain rule).

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  • $\begingroup$ Thanks very much! But how would that lead us to believe that the general solution looks like the one in the OP? $\endgroup$
    – dreamer
    Oct 11, 2015 at 12:11
  • $\begingroup$ Solving the constant-coefficient equation you generate gives a function that is a sum of terms of the form $e^{\lambda y}$. Substituting $x$ back in, $e^{\lambda y} = e^{\lambda \log{x}} = x^{\lambda}$. $\endgroup$
    – Chappers
    Oct 11, 2015 at 12:13
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Hint: see the Euler-Cauchy eqution https://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation

your solution should be $Y=Cx^{m}$ not $Y=Ce^{mx}$

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  • $\begingroup$ Thank you, but I'm not entirely sure what you mean, could you please explain? $\endgroup$
    – dreamer
    Oct 11, 2015 at 12:06

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