What is the relation between general solution of a differential equation and null space?

Question

When we try to solve a linear differential equation of 2nd order, to find the complete equation, we find the general solution and particular solution and add them to find the complete solution. To find general solution, we find the solution for [f(D)x]=0.
Similarly in Linear Algebra, to find null-space we do Ax=0. Are these equivalent? If not what is the relation between them?
It would be great if you can explain using an example.
Thanks!

Answer 1

To get the intuition from linear algebra, suppose that $A: \mathbb{R}^3 \to \mathbb{R}^3$ is a matrix of rank one. Then, the solution set $\ker A = \{x \in \mathbb{R}^3 \mid Ax = 0\}$ will be a plane through the origin. We can describe this plane by taking two spanning vectors $x_1, x_2 \in \ker A$, and then a general solution to the problem "find $x$ such that $Ax = 0$" is given by "all vectors of the form $t_1 x_1 + t_2 x_2$, for $t_1, t_2 \in \mathbb{R}^3$. This is the analogue of the homogeneous problem (indeed, an equation of the form $Ax = 0$ is called homogeneous). One of the most important properties of the homogeneous problem is that we can add any two solutions together to get another solution.

Now, what happens when we choose some $b \in \mathbb{R}^3$, and ask for solutions to the equation $Ax = b$? All of the solutions to this equation will form a plane parallel to the last one, but shifted away from the origin by $b$. Can we re-use our solutions for $Ax = 0$ in a meaningful way? It turns out the answer is yes: suppose a fairy comes along and gives you a single vector $y \in \mathbb{R}^3$ such that $Ay = b$, which she calls a particular solution. Then you can check that $A(y + t_1 x_1 + t_2 x_2) = b$, and so the general solution to the equation $Ax = b$ is $y + t_1x_1 + t_2 x_2$ for $t_1, t_2 \in \mathbb{R}^3$. You already know how to move around once you're inside the plane: this is just using $x_1, x_2$. But you do need the particular $y$ to get you to the shifted plane.

Now, let's move to differential equations. Let $V$ be the vector space of differentiable functions $\mathbb{R} \to \mathbb{R}$, and define the linear operator $D: V \to V$ by the formula $$Df = \frac{d^2f}{dx^2} - 9f$$ Then we would like to find vectors (i.e. functions) $f \in V$ such that $Df = 0$. Again, this is the kernel of the linear operator $D$, and so the set of solutions will form a linear subspace. We can quickly check that both $e^{3x}$ and $e^{-3x}$ are vectors solving the equation $Df = 0$, and so a linear combination $Ae^{3x} + Be^{-3x}$ will be a solution. (Some theorem about uniqueness of solutions to differential equations tells us that this is in fact, a completely general solution). So the solution set to $Df = 0$ looks like a plane through the origin (although planes in $V$ are kind of hard to visualise...).

You can probably fill in the rest about the particular solution now. You are asked to find all the functions $f$ solving $Df = g$, where $g(x)$ is some fixed function. A leprechaun comes along and gives you some function $p$ satisfying $Dp = g$, then a general solution to this new problem will be $p(x) + Ae^{3x} + Be^{-3x}$.