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This is supposed to be easily integrated to give the given solution, however I found myself easily puzzled on where to start

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  • $\begingroup$ Call $G=dF/dS$, so $d^2F/dS^2=dG/dS$. Your equation becomes a first order, separable one for $G$. $\endgroup$
    – GReyes
    Oct 3, 2021 at 0:46
  • $\begingroup$ It's like solving $ax+bx^2 = 0$. You can "factor" out an $x$ term to get $x(a+bx) = 0$. $\endgroup$
    – Toby Mak
    Oct 3, 2021 at 2:22

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That is actually a first order linear differential equation:

$$ \begin{align} 0&=\left(v-\mu S\right)\cdot\frac{dF}{dS}+\frac{\sigma^{2}}{2}S\cdot\frac{d^{2}F}{dS^{2}}\\ \\ &=\frac{2}{\sigma^{2}}\left(\frac{v}{S}-\mu\right)\cdot\frac{dF}{dS}+\frac{d}{dS}\frac{dF}{dS}\\ \\ &= \frac{2}{\sigma^{2}}\left(\frac{v}{S}-\mu\right)\cdot \exp\left({\int{\frac{2}{\sigma^{2}}\left(\frac{v}{S}-\mu\right)\phantom{x}dS}}\right)\cdot\frac{dF}{dS}+ \exp\left({\int{\frac{2}{\sigma^{2}}\left(\frac{v}{S}-\mu\right)\phantom{x}dS}}\right)\cdot\frac{d}{dS}\frac{dF}{dS}\\ \\ &=\frac{d}{dS}\left(\exp\left({\int{\frac{2}{\sigma^{2}}\left(\frac{v}{S}-\mu\right)\phantom{x}dS}}\right)\right)\cdot\frac{dF}{dS}+ \exp\left({\int{\frac{2}{\sigma^{2}}\left(\frac{v}{S}-\mu\right)\phantom{x}dS}}\right)\cdot\frac{d}{dS}\frac{dF}{dS}\\ \\ &=\frac{d}{dS}\left(\exp\left({\int{\frac{2}{\sigma^{2}}\left(\frac{v}{S}-\mu\right)\phantom{x}dS}}\right)\cdot\frac{dF}{dS}\right)\\ \\ &=\frac{d}{dS}\left(C_{1}\cdot S^{\frac{2v}{\sigma^{2}}}\cdot e^{-\frac{2\mu S}{\sigma^{2}}}\cdot\frac{dF}{dS}\right) \end{align} $$

Equivalent to:

$$ C_{2}=C_{1}\cdot S^{\frac{2v}{\sigma^{2}}}\cdot e^{-\frac{2\mu S}{\sigma^{2}}}\cdot\frac{dF}{dS}\\ \\ \text{or} \\ \\ A \cdot S^{-\frac{2v}{\sigma^{2}}}\cdot e^{\frac{2\mu S}{\sigma^{2}}}=\frac{dF}{dS} $$

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  • $\begingroup$ Clear and concise in exactly the form I want in the end! I guess I've been a bit rusty. Thanks! $\endgroup$ Oct 3, 2021 at 15:51

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