# Thread: Differential Equations

1. ## Differential Equations

Recall that a set of functions is linearly independent if their Wronskian is not identically 0. Calculate the Wronskian of the set to show that they are linearly independent. Write the differential equation for which they are solutions.

The set is:

$\displaystyle e^{\alpha x}, \ xe^{\alpha x}$ <---- I believe that these are not linearly independent, is that the case?

I can do the wronskian... What does it mean to write the differential equation? I'm not sure how to proceed.

2. I would do the Wronskian only to check: Let $\displaystyle a,b \in \mathbb{R}$ such that $\displaystyle ae^{\alpha x} +bxe^{\alpha x} \equiv 0$ then evaluating in$\displaystyle x=0$ gives $\displaystyle a=0$ and in $\displaystyle x=1$ gives $\displaystyle b=0$ so they are l.i.

As for the second question here's what I would do:

Clearly $\displaystyle g' - \alpha g =0$ is satisfied by $\displaystyle e^{ \alpha x}$. The second one is trickier let $\displaystyle a(x), b(x) \in C^0(\mathbb{R} )$ such that $\displaystyle b(x)f'(x)+a(x)f(x) \equiv 0$ then after some calculations we get $\displaystyle a(x)= \frac{ -b(x)(1+ \alpha x) }{ x}$ now $\displaystyle a(x) \notin C^0(\mathbb{R} )$ but we take $\displaystyle c(x)=x$ and $\displaystyle d(x)= 1+ \alpha x$ and we get $\displaystyle c(x)f'(x)+d(x)f(x) \equiv 0$ (this results from taking $\displaystyle b \equiv 1$, substituting $\displaystyle a,b$ in the equation and then multiplying by $\displaystyle x$ the whole thing). And with this equations we have a representation of $\displaystyle V:=span\{ e^{\alpha x}, xe^{\alpha x} \} \subseteq C^1(\mathbb{R} )$ and what I think you're asked is to try and find a second order (linear) dif. eq. such that $\displaystyle V$$\displaystyle \cap C^2(\mathbb{R} )$ is the set of solutions of said equation.

3. I appreciate that, my understanding in the subject is now a lot better because of that explanation, but you are right, the question "Write the differential equations for which they are solutions." means to find one differential equation that both of those satisfy, and it would surely be the one that has the solution set $\displaystyle V\cap C^2(\mathbb{R})$.

Can you get me started on finding that equation? I'll work on it and see what you think.

By the way, can you see what I've done wrong in the wronskian? I can't seem to get it to be identically zero...

$\displaystyle W = \left| \begin{array}{cc} e^{\alpha x} & xe^{\alpha x} \\ \alpha e^{\alpha x} & (e^{\alpha x} + \alpha x e^{\alpha x}) \end{array} \right|$

$\displaystyle W = e^{\alpha x}(e^{\alpha x} + \alpha x e^{\alpha x}) - \alpha x e^{2\alpha x} = e^{2\alpha x}$