Solving 2nd Order Diff Eq: X2y''-2xy'-4y=0

[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

 

[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!

 

[arildno]

EEh?

Have you even TRIED out your trial solution, franky?

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

 

[HallsofIvy]

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.)

 

[franky2727]

so then i get n²x²⁺ⁿ-4xⁿ=0

which does to n²x²-2xn-4=0 where can i go from here i don't 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

 

[franky2727]

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

 

[franky2727]

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

 

[Defennder]

Well you're told the trial solution already. Just use that and the answer for n. And note that your answer for n is incorrect. Check your calculations.

 

[HallsofIvy]

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

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

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.

 

[franky2727]

i had y=xⁿ y'=nx^n-1 y''=n²x^n-2

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

 

[Defennder]

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

 

[franky2727]

ahh got it now giving me n=-1 or +4 so what do i do with them now? sub them back into 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=
?

 

[HallsofIvy]

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 n^th 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.

 

[franky2727]

ahh ok thanks