$\displaystyle \varphi''+2\varphi'+\lambda\varphi=0$

$\displaystyle \varphi(0)=0$

$\displaystyle \varphi'(L)=0$

$\displaystyle \displaystyle m^2+2m+\lambda=0\Rightarrow m= -1\pm i\sqrt{\lambda-1}$

$\displaystyle \varphi=e^{-x}\left[ C_1\cos(x\sqrt{\lambda-1})+C_2\sin\left(x\sqrt{\lambda-1}) \right] $

$\displaystyle \varphi'=-e^{-x}\left[ C_1\cos(x\sqrt{\lambda-1})+C_2\sin\left(x\sqrt{\lambda-1}\right)+e^{-x}\left[C_2\sqrt{\lambda-1}\cos(x\sqrt{\lambda-1})-C_1\sqrt{\lambda-1}\sin\left(x\sqrt{\lambda-1}\right)\right]$

$\displaystyle \varphi_1(0): \ C_1=1, \ \ \varphi_1'(0): \ C_2=0$

$\displaystyle \varphi_1=e^{-x}\cos\left(x\sqrt{\lambda-1}\right)$

$\displaystyle \displaystyle\varphi_2(0): \ C_1=0, \ \ \varphi_2'(0): \ C_2=\frac{1}{\sqrt{\lambda-1}}$

$\displaystyle \displaystyle\varphi_2=e^{-x}\frac{\sin\left(x\sqrt{\lambda-1}\right)}{\sqrt{\lambda-1}}$

$\displaystyle \displaystyle\varphi(x)=e^{-x}\left[A\cos\left(x\sqrt{\lambda-1}\right)+B\frac{\sin\left(x\sqrt{\lambda-1}\right)}{\sqrt{\lambda-1}}\right]$

$\displaystyle \displaystyle\varphi(0): \ A=0, \ \ \varphi'(L): \ Be^{-L}\left[\cos\left(L\sqrt{\lambda-1}\right)-\frac{\sin\left(L\sqrt{\lambda-1}\right)}{\sqrt{\lambda-1}}\right]=0$

$\displaystyle \displaystyle\tan\left(L\sqrt{\lambda-1}\right)=\sqrt{\lambda-1}$

Is this correct and is this solvable?

Also, the Latex \right] isn't showing up in a few of the lines but it is there.