# Second order differential equation

## Homework Statement

Let V denote a vector space of twice differentiable functions on R. Define a linear map L on V by the formula:

$$L(u)=au''+bu'+cu$$

Suppose that $$u_{1},u_{2}$$ is a basis for the solution space of L(u)=0. Find a basis for the solution space of the fourth order equation L(L(u))=0. What can you say about the kernels of L and $$L^{2}$$?

## The Attempt at a Solution

Since u1,u2 is a basis for the solution space of L(u)=0, then the kernel of L is a plane spanned by $$u_{1},u_{2}$$.

Now, we want to find the basis for L(L(u))=0. Expanding L(L(u)), we get:

$$a(au''+bu'+c)''+b(au''+bu'+c)'+c(au''+bu'+c)=0$$, simplifying we get:

$$a^{2}u''''+2abu'''+(2ac+b^{2})u''+2bau'+c^{2}u=0$$

So the characteristic equation is:

$$a^{2}\lambda^{4}+2ab\lambda^{3}+(2ac+b^{2})\lambda^{2}+2bc\lambda+c^{2}=0$$

$$a^{2}\lambda^{4}+2ab\lambda^{3}+b^{2}\lambda^{2}+2ac\lambda^{2}+2bc\lambda+c^{2}=0$$

$$\lambda^{2}(a\lambda+b)^{2}+2c\lambda(a\lambda+b)+c^{2}=0$$

$$(\lambda(a\lambda+b)+c)^{2}=0$$

$$a\lambda^{2}+b\lambda+c=0$$

Now, since $$u_{1},u_{2}$$ is a basis for L(u)=0, then $$u_{1},u_{2}$$ must satisfies the characteristic equation $$a\lambda^{2}+b\lambda+c=0$$.

Thus, a basis for L(L(u))=0 is $$u_{1},u_{2}$$. The kernel of $$L^{2}$$ is also a plane spanned by $$u_{1},u_{2}$$.

If this is true, then the solution space of L(L(u))=0 is exactly the same as L(u)=0. I don't really understand the meaning of this, as in why is it the same? I think something is wrong but I am not sure.

Any help would be appreciated.

But further, you have already that L(u)= 0 only for vectors of the form $au_1+ bu_2$ for numbers a and b, so that if L(L(u))= 0 then you must have $L(u)= au_1+ bu_2$. Think of that as a non-homogeneous second order differential equation. Of course, $u_1$ and $u_2$ are already solutions to the homogeneous equation so you should try solutions like $u= xu_1(x)$ and $u= xu_2(x)$.