# Fundamental set of solutions

1. Feb 9, 2009

### d_b

For a second order linear differential homogeneous equation, if the two solution y1 and y2 is a multiple of one another. It means that it is linearly dependent which mean they can not form a fundamental set of solutions to second order differential homogeneous equation.

Am I correct?? or could it be any cases where y1 and y2 is a mulitple of one another and still can form a fundamental set of solutions.

Also if y1 and y2 are L.D is that mean wronskian equals zero????

2. Feb 10, 2009

### Emreth

You are correct, they cannot be multiples, they cannot be linearly dependent, as that means they are the same solution. (u=au1+bu1=cu1.)

3. Feb 10, 2009

### HallsofIvy

Staff Emeritus
And yes, if y1 and y2 are multiples of each other, then their Wronskian is equal to 0:
Specifically, if y1(t)= ay2(t) for some number a, then it is also true that y1'= ay2' so the Wronskian is
$$\left|\begin{array}{cc}y1 & y2 \\ y1' & y2'\end{array}\right|= \left|\begin{array}{cc} y1 & ay1 \\ y1' & ay1'\end{array}\right|= a(y1)(y1')- a(y1)(y1')= 0$$

4. Feb 10, 2009

### d_b

what about y2=U(t)y1 ? Isn't y1 and y2 is the set of fundamental solution?? why is wronskian is not equal to zero then???

5. Feb 10, 2009

### Emreth

those are not linearly dependent.If U was a constant instead of U(t), they would be.