Proof of Second Order ODE Theorem

[Dick]

Well you said that it's not in the span of x³ and |x³| so that means there is no linear combination of them which will give me c*x.

I said that. I think you should prove it. Once you've done that, does that save your theorem?

 

[STEMucator]

I said that. I think you should prove it. Once you've done that, does that save your theorem?

So supposing I want to find some linear combination of x³ and |x³| which gives me cx, I want to satisfy this relation :

c₁x³ + c₂|x³| = cx.

The only way this is true is if all the constants c₁, c₂ and c are all zero?

 

[Dick]

So supposing I want to find some linear combination of x³ and |x³| which gives me cx, I want to satisfy this relation :

c₁x³ + c₂|x³| = cx.

The only way this is true is if all the constants c₁, c₂ and c are all zero?

True. How would you prove that instead of just saying it's true?

 

[STEMucator]

True. How would you prove that instead of just saying it's true?

Hmmm okay so. I realize that I can take any straight line here including 0, so I'll use 0 since it's easy.

We have y₁ = x³ and y₂ = |x³| which are L.I solutions to our equation.

y'₂ = 3x² if x ≥ 0
y'₂ = -3x² if x < 0

Hence their Wronskian is zero everywhere for all x in our interval, which should mean they are linearly dependent, but this is not the case.

Let a and b be constants such that :

ax³ + b|x³| = 0

Then :
a(1)³ + b|1³| = a + b = 0
a(-1)³ + b(-(-1)³) = a - b = 0

So it must be the case that a = b = 0 and thus y₁ and y₂ must be linearly independent.

 

[Dick]

Hmmm okay so. I realize that I can take any straight line here including 0, so I'll use 0 since it's easy.

We have y₁ = x³ and y₂ = |x³| which are L.I solutions to our equation.

y'₂ = 3x² if x ≥ 0
y'₂ = -3x² if x < 0

Hence their Wronskian is zero everywhere for all x in our interval, which should mean they are linearly dependent, but this is not the case.

Let a and b be constants such that :

ax³ + b|x³| = 0

Then :
a(1)³ + b|1³| = a + b = 0
a(-1)³ + b(-(-1)³) = a - b = 0

So it must be the case that a = b = 0 and thus y₁ and y₂ must be linearly independent.

That's not very good. You already showed x^3 and |x^3| are linearly independent and you just repeated it. Now show me x^3 and |x^3| and x are linearly independent.

 

[STEMucator]

That's not very good. You already showed x^3 and |x^3| are linearly independent and you just repeated it. Now show me x^3 and |x^3| and x are linearly independent.

Well first I note that their Wronskian is zero everywhere, which means they are linearly dependent, which is also not the case.

Following a similar proof style to before I form the equation :

ax³ + b|x³| + cx = 0

and show that a = b = c = 0 is the only possible solution.

 

[Dick]

Well first I note that their Wronskian is zero everywhere, which means they are linearly dependent, which is also not the case.

Following a similar proof style to before I form the equation :

ax³ + b|x³| + cx = 0

and show that a = b = c = 0 is the only possible solution.

The wronskian doesn't have much to do with it. If you put x=1, x=(-1) and x=2 that should give you enough ammunition to show a=b=c=0, right?

 

[STEMucator]

The wronskian doesn't have much to do with it. If you put x=1, x=(-1) and x=2 that should give you enough ammunition to show a=b=c=0, right?

Yes, I didn't write it before out of laziness, but here it is :

ax³ + b|x³| + cx = 0

Then :

If x = 1, a + b + c = 0
If x = -1, a - b + c = 0
If x = 2, 8a + 8b + 2c = 0 → 4a + 4b + c = 0

System of 3 eqs with a determinant which is non - zero. Thus a = b = c = 0.