Given the following ODE. $$ y''-4\alpha y'+4\alpha^2y=0 $$ where $$ y(0)+y'(0)=0\\ y(1)+y'(1)=0 $$ where $y' = \frac{dy}{dx}$ and similarly $y'' = \frac{d^2y}{dx^2}$
Find the eigenvalue $\alpha$ and also the solution?
Eigen value.... Differential equations
0
$\begingroup$
real-analysis
boundary-value-problem
1 Answers
1
This equation has characteristic equation with repeated root $2\alpha$, so the general solution is of the form $y=e^{\alpha x}(Ax+B)$, with derivative $ y' = e^{\alpha x}(A+(Ax+B)\alpha). $ Then we want the function $$ y+y' = e^{\alpha x}((1+\alpha)(Ax+B)+A) $$ to have zeros at $x=0$ and $x=1$.
Putting $x=0$, $$ (1+\alpha)B+A = 0, $$ and putting $x=1$, $$(1+\alpha)(A+B)+A = 0,$$ which you can now solve to find $\alpha$ and $B$: subtracting gives $$ (1+\alpha)A = 0. $$ If $\alpha=-1$, we also find $A=0$ from the first equation, and vice versa. Hence $\alpha=-1$, $A=0$, and $B$ can have any value (as one would expect since rescaling $y$ makes no difference to the differential equation or boundary conditions.