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Solution of the associated homogeneous problem

  • Thread starter confused88
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  • #1
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Homework Statement


Please help me with this, I would really appreciate it:

Using the fact that y1 = x-1/2 cos x is a solution of the associated homogeneous problem, obtain the general solution of
x2 y'' + x y' + (x2 - 1/4)y = x3/2


The Attempt at a Solution



Well the first thing i did was divide it by x^2 to obtain:
y'' + 1/x y' + (1 - 1/4x2)y = x-1/2

Then let y'' + 1/x y' + (1 - 1/4x2)y = 0

A second solution will be given as
y2 = y1 [tex]\int e ^ \int P(x)\frac{}{} (y1)^2[/tex] eintegral of P(x) dx [tex]\div[/tex] (y1)^2

So here is where i have a problem, can someone please help me integrate

y2 = x-1/2 cos x [tex]\int[/tex] - x / ( x-1/2 cos x)^2

please help me..?
 

Answers and Replies

  • #2
22
0


Umm the first latex image generation is supposed to be an integral,
The second is supposed to be a fraction, so e^integral of P(x) divided by (y1)^2

And the third one is another integral.

Sorry I dunno what happened..

please help
 
  • #3
22
0


Umm the first latex image generation is supposed to be an integral,
The second is supposed to be a fraction, so e^integral of P(x) divided by (y1)^2

And the third one is another integral.

Sorry I dunno what happened..

please help me
 
  • #4
22
0


Umm the first latex image generation is supposed to be an integral,
The second is supposed to be a fraction, so e^integral of P(x) divided by (y1)^2

And the third one is another integral.

Sorry I dunno what happened..

please help me
 

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