Question

Differentials question:
For the Heat, wave, and laplace equations there are boundary conditions u1 and u2, if you alter the conditions by deriving them (u1*, u2*) then the series solution is just cosine equation with the same eigenvalue. But if you only derive one of the equations (mixed conditions) such as (u1*,u2) or (u1,u2*) the eigenvalues are altered but the same for both cases correct?

Answer 1

\[\lambda = \frac{ (2n-1)\pi }{ 4 }\] For both cases (u1*,u2) or (u1,u2*) the eigenvalue is just this correct?

Answer 2

\[\sum_{n->0}^{\infty} [\frac{ 2 }{ L }\int\limits_{0}^{L}(f(x)\sin \frac{ (2n-1)\pi }{ 4L }x)dx]* \sin \frac{ (2n-1)\pi }{ 4L }xe ^{\frac{ t(2n-1)^{2} \pi^{2} \alpha^{2} }{ 2L^{2} }} \]

I'm assuming the final solution will look like this?