1. ## \theta''+\lambda\theta=0

Let $\theta(x,\lambda)$ be the solution of

$\theta''+\lambda\theta=0$
$\theta(0)=1$
$\theta'(0)=0$

Use Green's formula to evaluate

$\displaystyle\int_0^L\theta^2(x,\lambda) \ dx$

and by specializing the result evaluate

$\displaystyle\int_0^L\cos^2\left(\frac{n\pi x}{L}\right) \ dx, \ \ \int_0^L\cos^2\left[\left(n+\frac{1}{2}\right)\frac{\pi x}{L}\right] \ dx$

where n is an integer.

Help!

Using Green's Formula, I obtain.

$\displaystyle\int_0^L\theta^2(x,\lambda) \ dx=\int_0^L[(A\theta)\bar{\theta}-\theta(A\bar{\theta}] \ dx=\theta'\bar{\theta}-\theta\bar{\theta}']_0^L=\theta'(L)\bar{\theta}(L)-\theta(L)\bar{\theta}'(L)$

where $\displaystyle A=-\frac{d^2\theta}{dx^2}$

Now what?

2. Sorry, but this question doesn't make a lot of sense as it's written. $\displaystyle \theta$ is a function of two variables, so this should be a PDE, but you've written it as an ODE. Which variable has the function been differentiated with respect to?

3. Originally Posted by Prove It
Sorry, but this question doesn't make a lot of sense as it's written. $\displaystyle \theta$ is a function of two variables, so this should be a PDE, but you've written it as an ODE. Which variable has the function been differentiated with respect to?
Lambda is the parameter the derivatives are wrt to x.

4. Are you given any restriction on $\displaystyle \lambda$?

5. Originally Posted by Prove It
Are you given any restriction on $\displaystyle \lambda$?
Nope. The full question with all information is posted.