Question

. Solve the following homogeneous differential equation: (x + 2y)dx − xdy = 0

Answer 1

$\left( {x + 2y} \right)dx - xdy = 0 \Rightarrow \frac{{dy}}{{dx}} = \frac{{x + 2y}}{x} \Rightarrow y' = 1 + 2\frac{y}{x}$

Let

$\frac{y}{x} = t \Rightarrow y = tx \Rightarrow y' = t'x + t$

Then

$t'x + t = 1 + 2t \Rightarrow x\frac{{dt}}{{dx}} = 1 + t \Rightarrow \frac{{dt}}{{1 + t}} = \frac{{dx}}{x} \Rightarrow \ln (1 + t) = \ln x + \ln C \Rightarrow 1 + t = Cx \Rightarrow 1 + \frac{y}{x} = Cx \Rightarrow y = x\left( {Cx - 1} \right) \Rightarrow y = C{x^2} - x$

Answer: $y = C{x^2} - x$