How to get general solution via Green's function?

[psie]

Homework Statement

Find a fundamental solution of the equation $x''-x=0$. Solve using this the equation $x''-x=e^t\sin{t}$.

Relevant Equations

The definition of a fundamental solution is given in theorem below.

I'll start with a characterization of the Green's function as a fundamental solution to a differential operator. This theorem is given in Ordinary Differential Equations by Andersson and Böiers.

Theorem. Let $$L(t,\lambda)=\lambda^n+a_{n-1}(t)\lambda^{n-1}+\ldots+a_1(t)\lambda+a_0(t)\quad\text{and }D=\frac{d}{dt}.\tag1$$ Denote by $E(t,\tau)$ the uniquely determined solution $u(t)$ of the initial value problem
\begin{align}
&L(t,D)u=0 \tag2\\
&u(\tau)=u'(\tau)=\ldots=u^{(n-2)}(\tau)=0,\quad u^{(n-1)}(\tau)=1. \tag3
\end{align}
Then
$$y(t)=\int_{t_0}^t E(t,\tau)g(\tau)d\tau\tag4$$
is the solution of the problem
\begin{align}
&L(t,D)y=g(t) \tag5\\
&y(t_0)=y'(t_0)=\ldots=y^{(n-1)}(t_0)=0. \tag6
\end{align}

$E(t,\tau)$ is known as the fundamental solution to the differential operator $L(t,D)$, also known as Green's function.

Now, consider the equation $x''-x=e^t\sin{t}$. The homogeneous solution is $x_h(t)=Ae^t+Be^{-t}$. To obtain $E(t,\tau)$, note that $x_h'(t)=Ae^t-Be^{-t}$. According to $(3)$, $x_h(\tau)=x_h'(\tau)=0$, so we have two equations and two unknowns. Solving this system, we obtain $A=\frac{e^{-\tau}}{2}$ and $B=-\frac{e^\tau}{2}$, so $$E(t,\tau)=\frac12(e^{t-\tau}-e^{\tau-t})=\sinh{(t-\tau)}.$$

Here is where I'm stuck. $(4)$ seems to depend on an initial value $t_0$, which I don't have. Moreover, I'm a little unsure whether or not $(4)$ is the general solution to $L(t,D)y=g(t)$ or simply a particular solution. I'm looking to obtain the general solution of $x''-x=e^t\sin{t}$ using Green's function, i.e. $E(t,\tau)$. Grateful for any help.

 

[renormalize]

Let's begin by rewriting your general theorem as it applies to your specific $2^{nd}$-order ODE, using the same equation numbers as in the theorem. Your Green's function $E\left(t,\tau\right)$ must satisfy:
$$\frac{d^{2}E\left(t,\tau\right)}{dt^{2}}-E\left(t,\tau\right)=0\tag{2}$$with
$$E\left(\tau,\tau\right)=0,\quad\left.\frac{dE\left(t,\tau\right)}{dt}\right|_{t=\tau}=1\tag{3}$$In your post, you solve this system for the unique result:$$E\left(t,\tau\right)=\sinh\left(t-\tau\right)$$which is clearly correct. Turning then to eq.$(4)$ of the theorem, you write:

Here is where I'm stuck. $(4)$ seems to depend on an initial value $t_{0}$, which I don't have.

True, we don't have a specific value for $t_{0}$, just like we don't have a specific value for $t$, so we just carry them both along. We make this manifest by altering the left of eq.$(4)$ to make it clear that the ultimate solution $x$ can depend on both of the variables $t$ and $t_{0}$:
$$x\left(t,t_{0}\right)=\intop_{t_{0}}^{t}E\left(t,\tau\right)g\left(\tau\right)d\tau=\intop_{t_{0}}^{t}\sinh\left(t-\tau\right)e^{\tau}\sin\left(\tau\right)d\tau\tag{4}$$So now you've arrived at the most tedious step: performing the integral on the right of $(4)$. It's elementary, in that it boils down to integrating products of exponentials, but the final answer is a sum over several terms involving trig functions and exponentials. (I "cheated" and used Wolfram Mathematica!). Once you've derived the explicit expression for $x(t,t_{0})$, you can directly verify that it does indeed solve the last two equations of the theorem:
$$\frac{d^{2}x\left(t,t_{0}\right)}{dt^{2}}-x\left(t,t_{0}\right)=g\left(t\right)=e^{t}\sin\left(t\right)\tag{5}$$and:
$$x\left(t_{0},t_{0}\right)=\left.\frac{dx\left(t,t_{0}\right)}{dt}\right|_{t=t_{0}}=0\tag{6}$$You next note:

I'm a little unsure whether or not $(4)$ is the general solution...

Well, the general solution $x$ to a $2^{nd}$-order ODE must involve two arbitrary constants of integration, which can be used to set arbitrary initial conditions on $x,x'$ at a particular point. So solution $(4)$ cannot be general since by eq.$(6)$ it has the very specific initial values $x=x'=0$ at $t=t_{0}$. But this is easily remedied: simply add to $x$ your arbitrary solution $x_{h}(t)$ of the homogeneous equation $x_{h}''-x_{h}=0$ to get the general solution $x_{g}$:$$x_{g}\left(t,t_{0}\right)\equiv x\left(t,t_{0}\right)+x_{h}\left(t\right)=x\left(t,t_{0}\right)+c_{1}e^{t}+c_{2}e^{-t}\qquad(c_{1},c_{2}\text{ arbitrary constants})$$$x_{g}$ still satisfies the inhomogeneous equation $(5)$, but the initial conditions $(6)$ generalize to:$$x_{g}\left(t_{0},t_{0}\right)=c_{1}e^{t_{0}}+c_{2}e^{-t_{0}},\quad\left.\frac{dx_{g}\left(t,t_{0}\right)}{dt}\right|_{t=t_{0}}=c_{1}e^{t_{0}}-c_{2}e^{-t_{0}}$$Evidently, by appropriately choosing $c_{1},c_{2}$ you have the freedom to set any desired initial conditions on $x_G$ and $x_G'$ at $t=t_{0}$, as is required for the general solution. Hence, you can express the general solution of your problem in terms of the Green's function as:$$x_{g}\left(t,t_{0}\right)=\intop_{t_{0}}^{t}\sinh\left(t-\tau\right)e^{\tau}\sin\left(\tau\right)d\tau+c_{1}e^{t}+c_{2}e^{-t}$$