Differential Equation Solutions By Substitutions

I'm really stuck on these problems...

1. x dx + (y - 2x) dy = 0

2. dy/dx = (y - x) / (y + x)


Thanks alot....

1)

x dx + (y - 2x) dy = 0
x dx = - (y - 2x) dy
x dx = (2x - y) dy
x / (2x - y) = dy/dx
x / y(2x/y - 1) = dy/dx
(x/y) (1 / (2x/y - 1)) = dy/dx

u = x/y . . . y = x/u

y ' = (u*1 - x * u')/u^2
y ' = (u - x * u')/u^2

Start Substituting:
u (1 / (2u - 1)) = (u - x * u')/u^2
u^3 = (u - x * u')(2u - 1)
u^3 = 2u^2 - u - 2u xu' + xu'
u^3 - 2u^2 + u = - 2u xu' + xu'
u^3 - 2u^2 + u = (- 2u + 1) x u'
1/x = (- 2u + 1) u' / (u^3 - 2u^2 + u)
1/x = (- 2u + 1) u' / u(u^2 - 2u + 1)
1/x = (- 2u + 1) u' / u(u - 1)^2 <----- we can start integrating partial faction

(- 2u + 1) / u(u - 1)^2 = [ A / u ] + [ B / (u - 1) ] + [ C / (u - 1)^2 ]
(- 2u + 1) = A (u - 1)^2 + B * u (u - 1) + C * u
(- 2u + 1) = A (u^2 - 2u + 1) + B * (u^2 - u) + Cu
(- 2u + 1) = Au^2 - 2uA + A + Bu^2 - uB + Cu

[ 1 = A ]
[ 0 = A + B ] u^2 -----> - A = B -----> B = -1
[ - 2 = - 2A - B + C ] u -----> -2 = -2(1) - -1 + C ---> -2 = -2 + 1 + C ----> -2 = -1 + C ----> C = -1

(- 2u + 1) / u(u - 1)^2 = [ 1 / u ] + [ -1 / (u - 1) ] + [ -1 / (u - 1)^2 ] ---> START INTEGRATING
∫ (- 2u + 1) / u(u - 1)^2 = ln( u ) - ln(u - 1) - (u - 1)^(-2+1)/(-2+1)
∫ (- 2u + 1) / u(u - 1)^2 = ln( u ) - ln(u - 1) - (u - 1)^(-1/(-1)
∫ (- 2u + 1) / u(u - 1)^2 = ln( u ) - ln(u - 1) + (1/(u - 1))

Back to original as
ln(x) + C = ln( u ) - ln(u - 1) + (1/(u - 1))
ln(x) + C = ln( x/y / (x/y - 1) ) + (1/(x/y - 1))
ln(x) + C = ln( x/y / (x/y - y/y) ) + (1/(x/y - y/y))
ln(x) + C = ln( x/y / (x - y)/y ) ) + (1/(x - y)/y )
ln(x) + C = ln( x / (x - y) ) ) + (y/(x - y) )

---------------------------

2)

dy/dx = (y - x) / (y + x)
dy/dx = x(y/x - 1) / x(y/x + 1)
dy/dx = (y/x - 1) / (y/x + 1)

u = y/x. . . . xu = y

y ' = u + xu'

u + xu' = (u - 1) / (u + 1)
(u + xu')(u + 1)= (u - 1)
( u^2 + u + xu u' + xu' )= u - 1
( u^2 + u + xu u' + xu' - u = - 1
u^2 + xu u' + xu' = - 1
u' x(u + 1 ) = - 1 - u^2
u' (u + 1 ) / (- 1 - u^2) = 1/x
- u' (u + 1 ) / (1 + u^2) = 1/x
u' (u + 1 ) / (1 + u^2) = - 1/x -----> start integrating


u' (u + 1 ) / (1 + u^2) = u' ( u / (1 + u^2) ) + ( 1 / (1 + u^2) ) -----> start integrating
u' (u + 1 ) / (1 + u^2) = (1/2) ln(1 + u^2) + tan^-1( u )

(1/2) ln(1 + u^2) + tan^-1( u ) = - ln(x) + C
(1/2) ln(1 + (y/x)^2) + tan^-1( y/x ) = - ln(x) + C

=========

keeeep asking :)