Find $y \in W_{2}^{1}[-1,1]$ s.t. $\forall x \in W_{2}^{1}[-1,1]$, $f(x)=\langle x, y \rangle$ Consider a Sobolev space $W_{2}^{1}[-1,1]$ with the following inner product:
$\langle x, y \rangle = \int_{-1}^{1} [x(t)y(t)+x^{\prime}(t)y^{\prime}(t)]dt$.
Let $f(x) = \int_{-1}^{1}e^{2t}x(t)dt$.  
I need to find $y \in$ the Sobolev space $W_{2}^{1}[-1,1]$ such that $\forall x \in W_{2}^{1}[-1,1]$, $f(x)=\langle x, y \rangle$.
However, I don't even know how to begin. I know that I need $\int_{-1}^{1} [x(t)y(t)+x^{\prime}(t)y^{\prime}(t)]dt = \int_{-1}^{1}e^{2t}x(t)dt$, but that's about where what looks familiar to me ends...
I've since been informed that what I need to do is find a weak solution of $-y^{\prime\prime} + y = e^{2t}$; however, I have no idea how one goes from the problem I have here to this, much simpler-looking, ODE. A detailed explanation as to where this comes from is needed. 
Also, I don't know what boundary conditions I'm supposed to use. I am not doing this for a Differential Equations course; I am doing it in a Functional Analysis course, where my exposure to differential equations in the past has been extremely limited (as in, haven't had a class in ODEs since 2003, and even then, it was very basic). Please explain this to me. Then, after I have my solution to the homogeneous form of the 2nd order linear ODE, I should be able to figure out the particular solution (I hope) using either variation of parameters or undetermined coefficients. 
 A: Okay, so let's write out the condition $f(x)=\langle x,y\rangle$...
$$\int_{-1}^1 x(t)y(t)+x'(t)y'(t)\,\mathrm{d}t=\int_{-1}^1 e^{2t}x(t)\,\mathrm{d}t.$$
Observe we may write
$$\begin{array}{ll} \displaystyle \int_{-1}^1 x'(t)y'(t)\,\mathrm{d}t & \displaystyle =\int_{-1}^1 \left[\frac{\mathrm{d}}{\mathrm{d}t}\big( x(t)y'(t)\big)-x(t)y''(t)\right]\mathrm{d}{t} \\ & \displaystyle =x(1)y'(1)-x(-1)y'(-1)-\int_{-1}^1 x(t)y''(t)\,\mathrm{d}t \end{array}$$
To get rid of the terms outside of the integral impose boundary conditions $y'(1)=y'(-1)=0$.
Then the original equation may be rewritten as
$$ \int_{-1}^1 \left[\color{Blue}{y(t)-y''(t)}\right]x(t)\,\mathrm{d}t=\int_{-1}^1 \color{Blue}{e^{2t}}\,x(t)\,\mathrm{d}t. $$
The only way this can be true for all functions $x(t)$ is if
$$-y''+y=e^{2t}. $$
With the ansatz $y=\alpha e^{2t}$ we find $y=-\frac{1}{3}e^{2t}$ is a particular solution, however this doesn't satisfy the correct boundary conditions. The homogeneous solutions are $e^t$ and $e^{-t}$, so we may write out the form $y=\alpha e^t+\beta e^{-t}-\frac{1}{3}e^{2t}$, in which case the boundary conditions you should be able to write out as linear equations to solve for in $\alpha$ and $\beta$.
