Can anybody help please

  • #1

Main Question or Discussion Point

I have got a general solution for the equation

xy'' - y' + 4x^3y = 0, x > 0

by converting to the normal version of the self adjoint form and solving with an auxiliary equation I have

y = Acos(2x) + Bsin(2x)

It then asks to select two independent solutions and verify the wronskian satisfies Abel's identity.

Can I simply set A = 0, B = 1 for y1 and A = 1, B = 0 for y2

I did this and I got 1 for the wronskian and kx for Abel's which doesn't prove it!!!

Can anybody tell me where I'm going wrong please
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,792
919
I have got a general solution for the equation

xy'' - y' + 4x^3y = 0, x > 0

by converting to the normal version of the self adjoint form and solving with an auxiliary equation I have

y = Acos(2x) + Bsin(2x)

It then asks to select two independent solutions and verify the wronskian satisfies Abel's identity.

Can I simply set A = 0, B = 1 for y1 and A = 1, B = 0 for y2

I did this and I got 1 for the wronskian and kx for Abel's which doesn't prove it!!!

Can anybody tell me where I'm going wrong please
If y= A cos(2x)+ B sin(2x), y'= -2A sin(2x)+ 2B cos(2x), and y"= -4Acos(2x)- 4Bsin(2x).
Putting those into the equation gives:
xy'' - y' + 4x^3y = -4Axcos(2)- 4Bxsin(2x)+ 2Asin(2x)- 2Bcos(2x)+ 4x^3Acos(2x)+ 4x^3Bsin(2x) which is definitely NOT equal to 0. y= Acos(2x)+ Bsin(2x) does NOT satisfy the equation.

Yes, dividing the equation by 1/x2 puts it into self-adjoint form but what did you do from there?
 
  • #3
I put it into the normal version of the self adjoint form.

Where d/dx{p(x)dy/dx}+q(x)y=0

t = integral of 1/p(x) = x^3/3

dt/dx = x^2

dy/dx = dy/dt.dt/dx = dy/dt.x^2

etc.

Then solved using the auxiliary equation. It probably is my working that is wrong, however the question I really needed answering is; if I get a solution to the differential equation, to get two linearly independent solutions from the general solution can I simply assign values to the constants.

Thank you for your response
 
  • #4
I get m=+-sqrt(-4)

=+-2i

y = exp(alpha*x){Acos(beta*x)+Bcos(beta*x)}

where alpha is real and beta is imaginary

therefore

alpha is = 0

beta is = 2

y = Acos(2x)+Bsin(2x)
 
  • #5
my values were for t not x, I forgot to substitute them back in

y = Acos(2t) + Bsin(2t) where t = x^2/2

it works now

Thanks for your help
 

Related Threads for: Can anybody help please

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
12
Views
2K
Top