Question

Solve (x+y)y'=x-y using homogeneous methods.

Answer 1

\[\Large -(x-y)+(x+y)y'=0 \]
Differential Equation in homogenous form:
\[\Large (-x+y)+(x+y)y'=0 \]

\[\Large M_y=1=N_x \Rightarrow \text{Exact} \]

Answer 2

Solution:
\[(-x+y) + (x+y)y' = 0\]

let y = ux
=> y' = u + xu'

\[-x + ux + (x + ux)(u + xu') = 0\]
\[x(u^2 + 2u - 1) + x^2(1+u)u' = 0\]
\[x(1+u)u' = 1-2u-u^2\]
\[\frac{ 1+u }{ 1-2u-u^2 }u' = \frac{ 1 }{ x }\]
\[\int\limits \frac{ -2(1+u) }{ 1-2u-u^2 }du = \int\limits \frac{ -2 }{ x }dx\]
\[\ln(1-2u-u^2) = -2\ln(x) + \ln(c)\]
\[\ln(1-2u-u^2) = \ln(cx ^{-2})\]
\[1-2\frac{ y }{ x } - \frac{ y^2 }{ x^2 } = \frac{ c }{ x^2 }\]
\[x^2-2xy-y^2 = c\]