Trying to solve a Boundary Value Problem where $y'' = (x^2)y$ and the initial conditions are $y(0)=0$ and $y(L)= 0$. I have a general solution to be $Y(b) = c_1 \cosh(xb) + c_2\sinh(xb)$. I know that $c_1=0$ now my question is when does $\sinh(xb) = 0$?
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1Are you sure about your general solution ? – Claude Leibovici Feb 06 '14 at 05:57
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I don't think your general solution is correct. – Daryl Feb 06 '14 at 07:01
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This is a nice classical trap for those who jump to solving the problem before having finished reading its formulation, to say nothing of trying to understand it. This equation cannot be solved in elementary functions! Even if you knew what to do with the parabolic cylinder functions, writing down a general solution for the equation $\,y''=x^2y\,$ would be the most stupid way to approach this rather trivial problem. You should have paid attention to the remarkable fact that the equation and boundary conditions are homogeneous, which implies that $y=0$ solves the problem. Are there any other solutions? The answer is No, being quite simple to verify. Just multiply the equation by $y$ and integrate it over the interval $(0,L)$. Then, integrating by parts, you will find $$ 0=-\int\limits_0^Ly''y\,dx+\int\limits_0^L x^2y^2\,dx=-y'y\Bigr|^L_0+\int\limits_0^L|y'|^2 dx +\int\limits_0^L x^2y^2\,dx=\int\limits_0^L\bigl(|y'|^2+x^2y^2\bigr)\,dx, $$ whence follows immediately $\,y'(x)=y(x)=0\;\forall\, x\in [0,L]$, since by definition, a solution should be either classical $y\in C^2[0,L]$, or weakly differentiable $y\in H^2(0,L)$.
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