Refreshing solving second order ODE

Question

I have a boundary value problem for the following differential equation $$\frac{d^2 v}{d \chi^2} = q^2 \left( v - C \right), \; 0<\chi<S \; and \;\; v(0)=v(S)=0 $$ where $q$ and $C$ are certain constants. As for now I got to this point $$\frac{d v^2}{d \chi^2} = q^2 \left( v - v_+ \right) \\ \frac{d}{d \chi}\left( \frac{d v}{d \chi} \right)^2 = 2 q^2 \left( v - v_+ \right) \\ \int \frac{d}{d \chi} \left( \frac{d v}{d \chi} \right)^2 = 2 q^2 \int \left( v - v_+ \right) \\ \left( \frac{d v}{d \chi} \right)^2 = 2 q^2 \left( \frac{v^2}{2} - v_+ + C_1 \right)= q^2 \left( {v^2} - 2 v_+ + 2 C_1 \right)\\ \frac{d v}{d \chi} = \pm q \sqrt{ {v^2} - 2 v_+ + 2 C_1 } $$ First question - as my boundary conditions are given on $v$ and not on the $\frac{d v}{d \chi}$ I have to now to integrate it by separation of variables, right?

Answer 1

Indeed the easiest way to go is to make the change of variable $u = (v - v_+)$, and then to find an answer of the sort $u = A \exp(q \chi) + B\exp(-q \chi)$. Adding the boundary conditions $v(0)=v(-s_+)=0$ we get $$ v_+ + A + B = 0\\ v_+ + A\exp(-s_+ q) + B\exp(s_+ q) $$ We find $A$ and $B$ and bingo we have a solution!

Answer 2

The first step should be $$ \dfrac{d}{d\chi}\left(\frac{dv}{d\chi}\right)^2 = 2q^2(v-C)\frac{dv}{d\chi} = 2q^2\dfrac{d}{d\chi}\left(\frac{v^2}{2}-Cv\right) $$ Then follow it through.