Question

Solve the differential equation (1 + yx) xdy + (1 − yx) ydx = 0

Answer 1

Answer on Question #68581, Math / Differential Equations

Solve the differential equation

$$(1 + yx) x dy + (1 - yx) y dx = 0$$

Solution

The trivial solutions

$$y(x) = 0 \text{ and } x = 0$$ $$\begin{array}{l} (1 + yx) x dy + (1 - yx) y dx = 0 \\ (1 + yx) x \frac{dy}{dx} + (1 - yx) y = 0 \\ \frac{dy}{dx} = y'(x) = \frac{y(yx - 1)}{x(1 + yx)} \\ t(x) = xy, t'(x) = (xy)' = y + xy' = \frac{t}{x} + x \frac{y(yx - 1)}{x(1 + yx)} = \\ = \frac{t}{x} + \frac{t(t - 1)}{x(1 + t)} = \frac{t + t^2 + t^2 - t}{x(1 + t)} = \frac{2t^2}{x(1 + t)} \\ \frac{dt}{dx} = \frac{2t^2}{x(1 + t)} \\ dt = \frac{2t^2}{x(1 + t)} dx \\ \end{array}$$

This is a separable ODE.

$$\frac{1 + t}{2t^2} dt = \frac{dx}{x}$$

Integrate both sides

$$\int \frac{1 + t}{2t^2} dt = \int \frac{dx}{x}$$ $$\begin{array}{l} \int \frac{1}{2t^2} dt + \int \frac{1}{2t} dt = \ln |x| - \frac{1}{2} \ln C \\ - \frac{1}{2t} + \frac{1}{2} \ln |t| = \ln |x| - \frac{1}{2} \ln C \\ \frac{1}{t} = \ln \left(\frac{C |t|}{x^2}\right) \\ \frac{1}{xy} = \ln \left(\frac{C |xy|}{x^2}\right) \\ xy \ln \left(C \left| \frac{y}{x} \right|\right) = 1 \\ \end{array}$$

Answer: $x = 0; y(x) = 0; xy \ln \left(C \left| \frac{y}{x} \right|\right) = 1.$

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