工程數學，請問圈起來的第8題怎麼解?

8.
(1) verify that y1 is a sol. of the eq.
pf :
y1 = x , which implies y' = 1 and y'' = 0
y'' - 2xy'/( 1 + x² ) + 2y/( 1 + x² )
= 0 - 2x/( 1 + x² ) + 2x/( 1 + x² )
= 0
Q.E.D.

(2)
find a 2nd sol. by reduction of order
Sol :
Let y2 = u * y1 = ux , then
y' = u' x + u*1 = u' x + u
y'' = u'' x + u' + u' = u'' x + 2u'

y'' - 2xy'/( 1+ x² ) + 2y/( 1 + x² ) = 0
( 1 + x² )y'' - 2xy' + 2y = 0
( 1 + x² )( u'' x + 2u' ) - 2x( u' x + u ) + 2ux = 0
u'' x + 2u' + u'' x³ + 2u' x² - 2u' x² - 2ux + 2ux = 0
u'' x + 2u' + u'' x³ = 0
( x³ + x )u'' + 2u' = 0

Let w = u' ( reduction of order )
( x³ + x )w' + 2w = 0
w'/w + 2/( x³ + x ) = 0
(1/w)(dw/dx) + 2/( x³ + x ) = 0
(1/w)dw + [ 2/( x³ + x ) ]dx = 0
∫ (1/w)dw + ∫ [ 2/( x³ + x ) ]dx = c1
ln｜w｜+ ∫ [ 2/x - 2x/(x²+1) ]dx = c1
ln｜w｜+ 2*ln｜x｜- ln｜x² + 1｜= c1
ln｜wx² / (x² + 1)｜= c1
wx² / (x² + 1) = ± e^c1 = c2
w = c2 * (x² + 1) / x² = c2 * ( 1 + 1/x² )

u' = w = c2 * ( 1 + 1/x² )
u = ∫ c2 * ( 1 + 1/x² ) dx = c2 * ∫ ( 1 + 1/x² ) dx = c2*( x - 1/x + c3 )
y2 = ux = c2*( x - 1/x + c3 ) x
We may take c2 = 1 and c3 = 0 because y2 is just one sol. of the eq.
Hence, y2 = ( x - 1/x ) x = x² - 1
Ans: y2(x) = x² - 1

(3) write the general sol.
Sol :
y
= c1*y1 + c2*y2
= c1*x + c2( x² - 1 )
Ans: y = c1*x + c2( x² - 1 )