- #1
schattenjaeger
- 178
- 0
y''-(2xy')/(1-x^2)+(2y)/(1-x^2)=0, we know that y1(x)=x is one solution, find the other
(starting at y2'' I'll let the multiplication and y1 be assumed)
so y2=u*y1
y2'=u'*y1+u*y1'
y2''=u''y+2u'y'+uy''
plug all that crap into the original and separate the variables and it simplifies too
u''/u'=-[2y'/y-2x/(1-x^2)]
which is what we expect it too, so I'm pretty sure I'm right up to here, then we integrate both sides
lnu'=-2lny+ln(1-x^2)+c1, plug x back into y, and exponentiate both sides
u'=x^-2(1-x^2)*C
so integrate again
u=-c(x^2+1)/x
y2=u*y = -(x^2+1)*x (I set c = x to cancel out that one x in the denominator)
but this is totally not what the book got
(starting at y2'' I'll let the multiplication and y1 be assumed)
so y2=u*y1
y2'=u'*y1+u*y1'
y2''=u''y+2u'y'+uy''
plug all that crap into the original and separate the variables and it simplifies too
u''/u'=-[2y'/y-2x/(1-x^2)]
which is what we expect it too, so I'm pretty sure I'm right up to here, then we integrate both sides
lnu'=-2lny+ln(1-x^2)+c1, plug x back into y, and exponentiate both sides
u'=x^-2(1-x^2)*C
so integrate again
u=-c(x^2+1)/x
y2=u*y = -(x^2+1)*x (I set c = x to cancel out that one x in the denominator)
but this is totally not what the book got