Solving (1+x^2)y''-2y'+2y=0: y=x & Beyond

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In summary: You can solve for v by taking the derivative with respect to x and using the Chain Rule. In summary, y= x is one solution of the equation (1+x2)y"- 2y'+ 2y= 0 and v= 2v/(x(1+x2)) is a separable differential equation.
  • #1
franky2727
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show that y=x is one solution of the equation (1+x2)y''-2y'+2y=0 and determine a second solution

im up to the IF which i get as eintegral((2/x0-(2x/1+x2) as does my answer sheet but this when comes out as =(x2)/(1+X2) i am getting confused with how to get here as i would of thought i would i end up with logs but how do i get logs from this with a 2 and a 2x in front? step by step please someone
 
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  • #2
franky2727 said:
eintegral((2/x0-(2x/1+x2) as does my answer sheet but this when comes out as =(x2)/(1+X2) i am getting confused with how to get here as i would of thought i would i end up with logs but how do i get logs from this with a 2 and a 2x in front?

eintegral((2/x-(2x/1+x2)

= e2logx - log(1+x2) + c

Then use elogy = y. :smile:
 
  • #3
Have you copied the problem correctly? You say that you are to show that y= x is a solution of (1+ x2)y"- 2y'+ 2y= 0 and that is not true: if y= x, then y'= 1 and y"= 0 so the equation becomes (1+ x2)(0)- 2(1)+ 2x which is NOT 0.
 
  • #4
sorry its -2xy' not -2y'
 
  • #5
does e2lnx - ln(1+x2) + c go to 2ln|x/1+x|
 
  • #6
franky2727 said:
does e2lnx - ln(1+x2) + c go to 2ln|x/1+x|

Nooo …

that's e2lnx e- ln(1+x2) ec,

= (elnx)2 e- ln(1+x2) ec :smile:
 
  • #7
Okay, you are given (1+ x2)y"- 2xy'+ 2y= 0 and asked first to show that y= x is a solution and then find another (independent) solution.

I see no reason to find an integrating factor (which apply to first order equations, not second order, anyway) here- you are not asked to solve the equation "from scratch". If y= x, then y'= 1 and y"= 0 so your equation becomes (1+x2)(0)- 2x(1)+ 2x= 0 which is true. Yes, y= x is a solution.

Now look for another solution of the form y= xu(x). Then y'= xu'+ u and y"= xu"+ 2u'. Putting that into the equation, (1+ x2)(xu"+ 2u')- 2x(xu'+ u)+ 2xu= (x+ x3)u"+ (2+ 2x2)u'- 2x2u'- 2xu+ 2xu= x(1+x2u"+ 2u'= 0.

Let v= u' so the equation becomes x(1+x2)v'+ 2v= 0. Then v'= 2v/(x(1+x2)) and that is a separable differential equation.
 

1. What is the meaning of "y=x & Beyond" in this equation?

The phrase "y=x & Beyond" refers to finding solutions to the equation beyond just the condition of y=x. This means exploring other possible values of y that satisfy the equation.

2. How do you solve this equation?

This equation is a second-order linear differential equation, which can be solved using standard techniques such as the method of undetermined coefficients or the method of variation of parameters. Alternatively, it can also be solved using power series methods.

3. What is the significance of (1+x^2) in this equation?

The term (1+x^2) is a coefficient that multiplies the second derivative of y in the equation. It affects the behavior and shape of the solutions to the equation.

4. Can this equation be solved using numerical methods?

Yes, this equation can also be solved using numerical methods such as Euler's method or Runge-Kutta methods. However, these methods may not provide exact solutions and are more computationally intensive.

5. What are the applications of this type of equation in science?

This type of equation is commonly used in physics and engineering to model systems that exhibit oscillatory behavior, such as a pendulum or a mass-spring system. It is also used in population dynamics and other areas of science to describe growth and decay phenomena.

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