MHB What Is the Basis for the Kernel of the Differential Operator \(D^4-2D^3-3D^2\)?

[karush]

ok I am new to this basis of kernel and tried to understand some other posts on this but they were not 101 enough

Find the basis for kernel of the differential operator $D:C^\infty\rightarrow C^\infty$,
$D^4-2D^3-3D^2$

this can be factored into

$D^2(D-3)(D+1)$

 

[Opalg]

ok I am new to this basis of kernel and tried to understand some other posts on this but they were not 101 enough

Find the basis for kernel of the differential operator $D:C^\infty\rightarrow C^\infty$,
$D^4-2D^3-3D^2$

this can be factored into

$D^2(D-3)(D+1)$

First, you need to work out what the question is asking for. The differential operator $D:C^\infty\rightarrow C^\infty$ take a smooth function $y = f(x)$ and differentiates it. The kernel of $D$ is the set of functions that it takes to zero, namely the constant functions. So a basis for the kernel of $D$ would consist of a single element, the constant function $1$.

But what the question is actually asking for is not the kernel of $D$, but the kernel of $D^4-2D^3-3D^2$. Using the factorisation $D^4-2D^3-3D^2 = D^2(D-3)(D+1)$ (and the fact that those factors commute with each other), what you need to do is to find the kernel of each separate factor.

For example, the kernel of $D-3$ consists of functions $y=f(x)$ such that $(D-3)y = 0$, in other words $\frac{dy}{dx} - 3y = 0$. The solution of that differential equation consists of multiples of $e^{3x}$, so a basis for the kernel of $D-3$ would be the function $e^{3x}$. Now do the same for the other two factors $D^2$ and $D+1$, to get a basis for $D^2(D-3)(D+1)$.