In my PDE course we write PDE's as $Lu = f$, where L is a linear operator, u is the solution, and f is the inhomogeneous part of the equation. I have to find the linear operator that satisfies $Lu=0$ for $u = 2t+e^{-t}$. This is the equivalent of finding the linear operator that has u in it's kernel. If $u = e^{t}$ it would be easy as L = $D_{t}-I$, but I don't know how to deal with the added factor of 2t.
Could anyone give a hint?
Here is a procedure for finding an operator $L$ with $Lu=0$: Consider $u,u',u'',\cdots$. Study their linear independence. If you can find a linear relation $\sum a_iu^{(i)}=0$ (a finite sum) then you can take $Lf=\sum a_i\frac {d^{i}} {dx^{i}}$ as yuor operator. This method works for both $u=2t+2^{-t}$ and $u=3t\cos (2t)$.
Suppose we have a pde of the form $\alpha^i(x)\dfrac{\partial u}{\partial x^i}=f(x)$.
Define the vector field $\textbf X=(\alpha^1(x),\dots,\alpha^n(x))$.
If the equation is homogeneous we know that the solutions of the equation belong to $Ker(L)$ (where $L$ is the differential operator given by $L:=\alpha^i(x)\dfrac{\partial }{\partial x^i}$), so the solutions are functions constant on the integral curves of $\textbf X$.
For the non-homogeneous case: consider the system
$$\dfrac{dx^1}{ds}=\alpha^1(x),\dots,\dfrac{d x^n}{ds}=\alpha^n(x),\dfrac{du}{ds}=f(x)\text { which gives }$$
$$\dfrac{dx^1}{\alpha^1(x)}=\dots=\dfrac{dx^n}{\alpha^n(x)}=\dfrac{du}{f(x)}.$$
In homogeneous case we change coordinates from $x^1,\dots,x^n$ to $s,\xi^1,\dots,\xi^n$ s.t. the integral curves of $\textbf X$ are given by $\xi^i=c_i\in\mathbb R$. In this case the functions that solve the equation are functions of variables $\xi^1,\dots\xi^{n-1}$. Now, since we have the non-zero term $f(x)$, we have to add
$$u=\int_0^s F(\sigma,\xi^1,\dots,\xi^{n-1})d\sigma + K(\xi^1,\dots,\xi^{n-1}).$$
So, we define the vector field $$\textbf Y:=\sum_{i=1}^n \alpha^i(x)\dfrac{\partial}{\partial x^i}+f(x)\dfrac{\partial}{\partial u}=\textbf X+f(x)\dfrac{\partial}{\partial u}$$
and we find its integral curves.
Example: $tu_t+xu_x=xt$. We want to find integral curves of $\textbf Y=t\dfrac{\partial }{\partial t}+x\dfrac{\partial}{\partial x}+xt\dfrac{\partial}{\partial u}$.
Consider the system $\dot\gamma(s)=\textbf Y(\gamma(s))$, where $\gamma$ has components $\gamma^1=t(s),\gamma^2=x(s),\gamma^3=u(s)$.
From $\dfrac{dt}{t}=\dfrac{dx}{x}=\dfrac{du}{xt}$ we get $\xi=\dfrac{x}{t}$.
Then $\dfrac{dt}{t}=\dfrac{du}{xt}\implies dt=\dfrac{du}{x}\implies xdt=du\implies\xi tdt= du$, which gives $u-K=\xi\dfrac{t^2}{2}$.
Finally $u=K(\xi)+\xi\dfrac{t^2}{2}$.
