I have the following problem:
I have a second-order linear ODE $$\frac{1}{2}\sigma^2a^2f''(x)+\mu a \cdot f'(x)-\rho f(x)=0$$ My exercise is to show that $$f(x)=c_1\left(x\frac{\rho}{\gamma}+c_2 \right)^\gamma, \qquad \text{where }\gamma:=\frac{2\sigma^2\rho}{\mu^2+2\sigma^2\rho}.$$
My steps: First, I've found the interior maximizer for $a$, i.e. $$\frac{\partial}{\partial a}\left(\frac{1}{2}\sigma^2a^2f''(x)+\mu af'(x)-\rho f(x)\right)=\sigma^2\cdot a\cdot f''(x)+\mu f'(x)=0 \Longleftrightarrow a=-\frac{\mu}{\sigma^2}\frac{f'(x)}{f''(x)}.$$
Next, I put the maximizer into the linear ODE and get \begin{align*} 0 &= \frac{1}{2}\sigma^2a^2f''(x)+\mu a \cdot f'(x)-\rho f(x) &\Leftrightarrow\\ \rho f(x) &=\frac{1}{2}\cdot\frac{\mu^2}{\sigma^2}\frac{[f'(x)]^2}{f''(x)}-\frac{\mu^2}{\sigma^2}\frac{[f'(x)]^2}{f''(x)} &\Leftrightarrow\\ \rho f(x) &= -\frac{1}{2}\frac{\mu^2}{\sigma^2}\frac{f'(x)}{f''(x)}\cdot f'(x) &\Leftrightarrow\\ -\frac{2\rho\sigma^2}{\mu^2} &= \frac{f'(x)\cdot f'(x)}{f''(x)\cdot f(x)} \end{align*} How can I find the general solution of $f(x)$?
EDIT:
By using the rules of logarithm, I get: $$\frac{f'(x)}{f(x)}=[\ln(f(x))]'\quad\text{and}\quad \frac{f'(x)}{f''(x)}=c\Longleftrightarrow \frac{1}{c}=\ln[f'(x)]'\Leftrightarrow c = \frac{1}{\ln[f'(x)]'}.$$ Then, since $x\in[0,1]$, I get \begin{align*} -\frac{2\rho\sigma^2}{\mu^2} &= \frac{\ln(f(x)'}{\ln(f'(x))'} &\Leftrightarrow\\ -\frac{2\rho\sigma^2}{\mu^2} &= [\ln(f(x))-\ln(f'(x))]' &\Leftrightarrow\\ -\frac{2\rho\sigma^2}{\mu^2}\int_0^1 dx &= \ln(f(x))-\ln(f'(x)) &\Leftrightarrow\\ -\frac{2\rho\sigma^2}{\mu^2} &=\ln\left(\frac{f(x)}{f'(x)}\right). \end{align*} Let $\delta:=\frac{2\rho\sigma^2}{\mu^2}$ be defined. Then, by using exponential's rule we get \begin{align*} \exp(-\delta) &= \frac{f(x)}{f'(x)} &\Leftrightarrow\\ f'(x)-\exp(\delta)f(x)=0. \end{align*} Now I have a stupid question: How can I solve it?
we have $$\int \frac{f'(x)}{f(x)}dx$$ the soltion is $$\ln(|f(x)|)+C$$ if $f(x)\ne 0$ for all real $x$