Linear Independence of Solutions in Second Order ODEs

[Clausius2]

Assume the next differential LINEAR second order equation:

$$w''+\frac{4}{x}w'+\frac{4}{x^4}w=0$$

So I thought: OK, I need two independent solutions $$w_1$$ and $$w_2$$, because the space of solutions is of dimension two.

Then the professor gave us a solution:

$$w=sen(2/x)-(2/x)cos(2/x)$$

and I thought: Ok, the solution he's giving us is composed by two linearly independent functions (because doing the wronskian it does not becames zero anywhere), and therefore each function must be solution of the differential equation..

Is this last bolded statement true for any second linear ODE??.

 

[arildno]

Well, just try it out, and you''ll find that $$\sin(\frac{2}{x})$$ is NOT a solution of your diff.eq.

 

[Clausius2]

Well, just try it out, and you''ll find that $$\sin(\frac{2}{x})$$ is NOT a solution of your diff.eq.

Of course I tried it, and I realized of it. But I really thought that in a set of solutions $$c_1w_1+c_2w_2$$ in which w1 and w2 are linearly independant functions, w1 and w2 must satisfy EACH ONE the ODE.

 

[arildno]

Of course I tried it, and I realized of it. But I really thought that in a set of solutions $$c_1w_1+c_2w_2$$ in which w1 and w2 are linearly independant functions, w1 and w2 must satisfy EACH ONE the ODE.

No, no, Clausius2.

Suppose a solution S of a diff. eq can be written in terms of two functions, S=f+g.
Let us assume that the diff.eq is linear and homogenous.

Let L be our differential operator, so that we have LS=0.
From this, we have Lf+Lg=0

It by no means now follows that we must have Lf=Lg=0!
We can, for example let Lf=H=-Lg
where H is some function.
But then we have for example that Lf=H, that is f is NOT a solution to your original differential equation, Ly=0 (for which S is a solution).

 

[saltydog]

Any linear combination of solutions of a linear homogeneous DE is also a solution. However, not all solutions given, are linear combinations of solutions. Your solution above does not appear to be a linear combination of solutions. Thus I would suppose there is another solution such that:

$$w(x)=A[Sin(2/x)-2/xCos(2/x)]+Bg(x)$$


Me, I'd multiply by $x^4$, review power series, and figure out what that other solution is . . . just might.

Edit: I'd try reduction of order first.

 

[Clausius2]

OK. Thanks very much both of you. I've understood the point.