Question

dy/dx=(y-x)/(x-4y)

Answer 1

Answer on Question #74061 – Math – Differential Equations

Question

$$\frac{dy}{dx} = \frac{y - x}{x - 4y}$$

Solution

Transforming the right side we get:

$$\frac{y - x}{x - 4y} = \frac{\frac{y}{x} - 1}{1 - 4\frac{y}{x}}$$

Now let $\frac{y}{x} = t$, then $y = tx$, and $\frac{dy}{dx} = \frac{dt}{dx} x + t$. So we have:

$$\frac{dt}{dx} x + t = \frac{t - 1}{1 - 4t}$$ $$\frac{dt}{dx} x = \frac{t - 1}{1 - 4t} - t = \frac{t - 1}{1 - 4t} - \frac{t(1 - 4t)}{1 - 4t} = \frac{t - 1 - t + 4t^2}{1 - 4t} = \frac{4t^2 - 1}{1 - 4t}$$ $$\frac{dx}{x} = \frac{1 - 4t}{4t^2 - 1} dt = \frac{dt}{(2t)^2 - 1^2} - \frac{t}{t^2 - \frac{1}{4}} dt$$

Integrating: $\int \frac{dx}{x} = \int \frac{dt}{(2t)^2 - 1^2} - \int \frac{tdt}{t^2 - \frac{1}{4}}$

We've got: $\ln |x| = \frac{1}{2}\ln \left|\frac{2t - 1}{2t + 1}\right| - \frac{1}{2}\ln \left|t^2 - \frac{1}{4}\right| + \ln |C|$

Thus: $x = C \sqrt{\frac{2t - 1}{(2t + 1)(t^2 - \frac{1}{4})}} = C \sqrt{\frac{t - \frac{1}{2}}{(t + \frac{1}{2})(t - \frac{1}{2})(t + \frac{1}{2})}} = C \frac{1}{t + \frac{1}{2}}$

Remembering that $t = \frac{y}{x}$, we have: $\frac{y}{x} + \frac{1}{2} = \frac{C}{x}$, and $y = C - \frac{x}{2}$

Answer: $y = C - \frac{x}{2}$

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