# Where to start?

1. Aug 17, 2008

### franky2727

not a clue where to begin with this one or what method to use can someone start me off? its only revision not homework

the second order differential equation X2y''-2xy'-4y=0

has non-constant coefficients.find its general solution by looking for a sollution in the form Y=Xn and obtaining a quadratic equation for n

2. Aug 17, 2008

### Tomtom

Well, I guess you are to solve for the zeroes of that equation. What you've got there is an ax^2+bx+c=0, where a=y'', b=2y' and c=-4y.

Edit: Alrighty, I'm wrong ;) Cheers to HallsofIvy!

Last edited: Aug 17, 2008
3. Aug 17, 2008

### arildno

EEh?

Have you even TRIED out your trial solution, franky?

The characteristic equation you'll get on n is perfectly solvable.

4. Aug 17, 2008

### HallsofIvy

Staff Emeritus
That is what is referred to as an "Euler-type" or "Equiponential" equation. "Where to start" is by doing exactly what you are told to do! If y= xn, what are y' and y"? What happens if you put them into this different equation?

(No, Tomtom, that is not what is meant.)

5. Aug 19, 2008

### franky2727

so then i get n2x2+n-4xn=0

which does to n2x2-2xn-4=0 where can i go from here i dont see how this has helped? i still have two unknown constants or do i just sub in a value for n now?
and also if n=0 then -4=0

6. Aug 19, 2008

### franky2727

scratch that i've cocked it up ill just work it out again

7. Aug 19, 2008

### franky2727

am i right to get n=1+or- root 20 over 2? then which one do i sub back in to get the general?

8. Aug 19, 2008

### Defennder

Last edited: Aug 19, 2008
9. Aug 19, 2008

### HallsofIvy

Staff Emeritus
Since you are not showing what you are doing, all I can say is "absolutely not!". That is not even close to correct.
Please show exactly what characteristic equation (the equation for "n") you got and how you attempted to solve it.

10. Aug 19, 2008

### franky2727

subbing in and dividing through by xn i get n2-2n-4=0 giving me 1+ or - root 20 over 2

11. Aug 19, 2008

### Defennder

Your quadratic equation is incorrect. Note that $$y'' \neq n^2x^{n-2}$$.

12. Aug 19, 2008

### franky2727

ahh got it now giving me n=-1 or +4 so what do i do with them now? sub them back in to y y' and y''? giving me one equation that goes to 0=0 when using n=4 and the other of x+2x^-1 -8x^-3=0 or do i just sub in y' and y'' leaving y as it is and therefore finding an equation (2 equations) in the form of y=
?

13. Aug 19, 2008

### HallsofIvy

Staff Emeritus
Remember that you were told to try a solution of the form xn. You found that such a solution works if n= -1 or n= 4. In other words, x-1 and x4 are solutions to the differential equation. You should also know that the solutions to any nth order linear homogeneous differential equation for a "vector space of dimension n" which simply means that the general solution can be written as a linear combination of n independent solutions. Here, x-1 and x4 are definitely independent so any solution can be written as a linear combination of them: of the form Cx-1+ Dx4 where C and D can be any constants.

14. Aug 19, 2008

### franky2727

ahh ok thanks