Re: differential equations help

[confusedM]

I'm having a couple problems with my diff eq assignment if anyone can help me. The problem is 1. (a) show that the diff eq. for y=c1u1(x) + c2u2(x), where u1(x) and u2(x) are functions which are at least twice differentiable, may be written in determinant form as
| y u1(x) u2(x) |
| y u1'(x) u2'(x) | = 0
| y u1''(x) u2''(x)|

b) what happens in the case where

w = | u1(x) u2(x) | = 0
| u1'(x) u2'(x)|

 

[HallsofIvy]

I'm having a couple problems with my diff eq assignment if anyone can help me. The problem is 1. (a) show that the diff eq. for y=c1u1(x) + c2u2(x), where u1(x) and u2(x) are functions which are at least twice differentiable, may be written in determinant form as
| y u1(x) u2(x) |
| y u1'(x) u2'(x) | = 0
| y u1''(x) u2''(x)|

b) what happens in the case where

w = | u1(x) u2(x) | = 0
| u1'(x) u2'(x)|

In that case, the two function, u1(x) and u2(x), are not independent. You need two independent solutions to write the general solution to a second order equation.

 

[confusedM]

I understand that if the determinant is zero, the equations are linearly dependent, but I don't know how to go about doing the proof required to answer the problem.

 

[Defennder]

The question didn't ask you to prove it. It just asks "what happens in that case". Wikipedia happens to have the proof:
http://en.wikipedia.org/wiki/Wronskian#Proof:_Wronskian_and_linear_independence

 

[confusedM]

That's part B... part A though asks for proof of the determinant of the three equations equaling zero...that's the part I'm really having issues with.

 

[Defennder]

Well you didn't provide the DE, so I assume it's a 2nd order ODE. Is the first column of the matrix supposed to read y,y',y'' from top to bottom instead? HallsOfIvy has said you need 2 linearly independent solutions for a second order ODE, so what can you say if the determinant of that is zero?

 

[confusedM]

yes it is supposed to read y, y', y"... the problem is as it is written in the book... it didn't say that it was a 2nd order ODE.
Clearly the original differential eq. and it's two derivatives are linearly dependent since the determinant is zero, but I don't understand what I'm exactly supposed to be showing

 

[Defennder]

This can be verified directly. You are given y=c1u1+c2u2. Find y' and y'' and evaluate the determinant. No proof is needed because none is asked for.

 

[HallsofIvy]

Yes, Defennder is right. I had overlooked A and answered B.

If y= c1u1+ c2u2, then
\left|\begin{array}{ccc} y & u & v \\ y' & u' & v' \\ y" & u" & v" \end{array}\right|= \left|\begin{array}{ccc} c1u+ c2v & u & v \\ c1u1'+ c2u2' & u' & v' \\ c1u1"+ c2u2" & u" & v" \end{array}\right|

If you know a little linear algebra, and properties of determinants, the fact that the first column is a linear combination of the second and third columns, that tells you all you need. If you don't recognise that, go ahead and do the calculations.