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

[franky2727]

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 e^{integral((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

 

[tiny-tim]

e^{integral((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?

e^{integral((2/x-(2x/1+x2)}

= e^{2logx - log(1+x2) + c}

Then use e^logy = y.

 

[HallsofIvy]

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.

 

[franky2727]

sorry its -2xy' not -2y'

 

[franky2727]

does e^{2lnx - ln(1+x2) + c} go to 2ln|x/1+x|

 

[tiny-tim]

does e^{2lnx - ln(1+x2) + c} go to 2ln|x/1+x|

Nooo …

that's e^2lnx e^{- ln(1+x2)} e^c,

= (e^lnx)² e^{- ln(1+x2)} e^c

 

[HallsofIvy]

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.