Solve $f'' - \frac{iw}{\upsilon}f=0$ We are given $u=u(y,t)$ and $$\frac{\partial u }{\partial t}= \upsilon \frac{\partial ^2 u }{\partial y^2}$$where $\upsilon$ is the viscosity. Look for solutions of the form $u=Re\\{Ue^{iwt}f(y)\\}$ where $U$ is a constant. Sub this in the above gives $$\frac{d^2 f}{dy^2} - \frac{iw}{\upsilon}f=0$$ * * * Please can someone show how to get the general solution to this because the one I have in my notes is not making sense to me. It ends up being $$f=Ae^{-(1+i)ky}+Be^{(1+i)ky}$$ where $k=\sqrt{w}/\sqrt{2 \upsilon}$. Don't know where the $2$ comes from and the $(1+i)$.

$$\frac{d^2 f}{dy^2} - \frac{iw}{\upsilon}f=0$$ Looking for solutions on the form $e^{ay}$ requires : $$\frac{d^2 e^{ay}}{dy^2} - \frac{iw}{\upsilon}e^{ay}=0 \quad\to\quad a^2-\frac{iw}{\upsilon}=0 \quad\to\quad a=\pm\sqrt{\frac{w}{\upsilon}}\:i^{1/2}$$

$i=e^{i\frac{\pi}{2}}\quad\to\quad i^{1/2}=\left(e^{i\frac{\pi}{2}}\right)^{1/2}=e^{i\frac{\pi}{4}}=\cos(\frac{\pi}{4})+i \cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}}(1+i)$ $$a=\pm\sqrt{\frac{w}{\upsilon}}\:\frac{1}{\sqrt{2}}(1+i)=\pm\sqrt{\frac{w}{2\upsilon}}\:(1+i)=\pm k(1+i)$$ where $k=\sqrt{\frac{w}{2\upsilon}}$

The two basic solutions are : $e^{ay}=e^{\pm k(1+i)y}$