Analyitical solution to function and second deritive of function

[munkifisht]

I have the following (fairly simple) boundary value problem and I am trying to find an analytical solution to it, but for the life of me it's not working out. This is part of a larger thing where I'm trying to understand FEM and BVPs. Essentially this is a diffusion reaction problem.

My problem is I have the following (π=pi)

v-kv''=f(x)=sin(πx), x is between 1 and 2
v@x=0 = 0 and v@x=1 = 0

I have some plots but they are not matching my analytical solution (I think my brain has just broken tonight), but if anyone can steer me right I'd be grateful.

 

[munkifisht]

I should also have mentioned k is a constant

 

[HallsofIvy]

The differential equation is -kv''+ v= sin(\pi x)? That's a linear equation with constant coefficents. Its characteristic equation is -kr^2+ 1= 0 so that r= \pm 1 and the general solution is C_1e^x+ C_2e^{-x} (it could also be written as C_1cosh(x)+ C_2sinh(x)).

Look for a specific solution to the entire equation of the form Asin(\pi x)+ B cos(\pi x).

But I am concerned about the information that the differential equation only holds between x= 1 and x= 2, so that the previous solution is valid only between x= 1 and x= 2, while we are given the value of v at x= 0. Not knowing what v is like between 0 and 1, it is impossible to use that information. I suggest you recheck that- either the d.e. holds between 0 and 1 or the boundary values are given at 1 and 2.

 

[munkifisht]

#sorry, mistake on my part, yes, the equation holds between 0 and 1. Thanks for that, so rusty on differential calculus