If $\frac{d^2v}{dx^2}=a^{2}v$ and $v=v_0$ at $x=0$ and $v=0$ at $x=l$, then find the value of $v$

Question

Using $D$-Operator method, in the answer, it is given that $v=v_0\frac{\sinh a(l-x)}{\sinh l}$. But after solving it, i got $v_0=c_1+c_2$ and $c_1e^{al}+c_2e^{-al}$ after solving $m^{2}=a^{2}\implies m=\pm a$. But even after considering all the factors and solving these equations, i am still not getting the answer as given in the book. Please help $$c_1=\frac{v_0}{1-e^{2al}} \\ c_2=\frac{-v_0e^{2al}}{1-e^{2al}}$$

Answer 1

With the symmetric exponents it is sometimes helpful to combine them from the start into the odd and even parts of the exponential, as per $$ e^{\pm ax}=\cosh(ax)\pm\sinh(ax). $$ The only term with a root on the right side is the hyperbolic sine and has it at $x=0$, which is not helpful with the boundary conditions. However, adding a constant in the exponent only changes the coefficients, so that one can write the general solution also as $$ v(x)=c_1\cosh(a(x-l))+c_2\sinh(a(x-l)). $$ From the second boundary condition it now directly follows that $c_1=0$ and inserted into the first $$ v_0=v(0)=c_2\sinh(-al)=-c_2\sinh(al). $$