# IF/log question

1. Aug 16, 2008

### 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 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

2. Aug 16, 2008

### tiny-tim

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

= e2logx - log(1+x2) + c

Then use elogy = y.

3. Aug 17, 2008

### HallsofIvy

Staff Emeritus
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. Aug 17, 2008

### franky2727

sorry its -2xy' not -2y'

5. Aug 17, 2008

### franky2727

does e2lnx - ln(1+x2) + c go to 2ln|x/1+x|

6. Aug 17, 2008

### tiny-tim

Nooo …

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

= (elnx)2 e- ln(1+x2) ec

7. Aug 17, 2008

### HallsofIvy

Staff Emeritus
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.