TUT0102 QUIZ1

[Ken Chen]

Q：dy/dx  = (x+3y)/(x-y)

dy/dx = (1+(y/x)/(1-y/x)
y/x = u
y = ux
dy/dx = (d(ux))/dx = xdu/dx + udx/dx = (1+3u)/(1-u)
xdu/dx = (1+2u+u^2)/(1-u)
1-u〖(1+u)〗^2du = ∫▒〖1/x dx〗             
-∫▒〖(u+1-2)/〖(1+u)〗^2  du〗 = ∫▒〖1/x dx〗
-∫▒〖(u+1)/〖(1+u)〗^2  du〗 - ∫▒〖2/〖(1+u)〗^2  du〗 = ∫▒〖1/x dx〗
-ln⁡(1+u) - 2/(1+u) + C = ln⁡x       
-ln⁡(1+y/x)  - 2/(1+y/x) + C = ln⁡x
C - 2x/(x+y) = ln⁡x (1+y/x)
C - 2x/(x+y) =  ln⁡〖(x〗+y)
2x/(x+y) + ln⁡〖(x〗+y)=C
x + y = Ce^((-2x)/(x+y))