(X-y)dx + (3x + y)dy =0

Question

Differential equation problem

Answer 1

Solve: (x-y)dx + (3x + y)dy =0

Rearranging,

dx/dy = -(3x + y)/(x – y)

dx/dy = -(3v + 1)/(v – 1), where v = x/y  ---------------------------------- (1)

Differentiating x = vy wrt y,

dx/dy = v + y.dv/dy   ----------------------------------------------------------- (2)

Substituting for dx/dy from (2) into (1),

v + y.dv/dy = -(3v + 1)/(v – 1)

y.dv/dy = -(3v + 1)/(v – 1) – v(v – 1)/(v – 1) = {-3v – 1 – v^2 + v}/(v – 1)

y.dv/dy = {– v^2 – 2v – 1}/(v – 1)

y.dv/dy = (1 + v)^2/(1 – v)

{(1 – v)/(1 + v)^2}.dv = (1/y).dy

Integrating both sides independently,

-ln(1 + v) – 2/(1 + v) = ln(y) + ln(A)

ln(Ay(1 + v)) = -2/(1 + v)

y = B/(1 + v)*exp(-2/(1 + v))

Substituting back in for v = x/y

y = B/(1 + x/y)*exp(-2/(1 + x/y))

1 = B/(x + y)*exp(-2y/(x + y))

x + y = B*exp(-2y/(x + y))

If you differentiate the above expression, and manipulate it, you will be able to get back to the original DE.