Question

(x^2+y^2)dx − 2xydy=0

−x^2−y^2=cx
−x^2+y^2=cx
x^2+y^2=cx
x^2−y^2=cx

Answer 1

Answer on Question # 79329 – Math – Differential Equations

Question

Solve the differential equation:

$$(x^2 + y^2) \, dx - 2xy \, dy = 0$$

Solution

$$(x^2 + y^2) \, dx - 2xy \, dy = 0$$

Or, $(x^2 + y^2) - 2xy \frac{dy}{dx} = 0$

Or, $\frac{dy}{dx} = \frac{(x^2 + y^2)}{2xy}$ ... (1)

Let $y = vx$ ... (2) [v is the function of x]

Now differentiate equation (2) with respect to x and we get,

$$\frac{dy}{dx} = v + x \frac{dv}{dx}$$

Now, put the values of $\frac{dy}{dx}$ and $y = vx$ in equation (1), we get,

$$v + x \frac{dv}{dx} = \frac{(1 + v^2)}{2v}$$

or, $x \frac{dv}{dx} = \frac{(1 + v^2)}{2v} - v = \frac{(1 - v^2)}{2v}$

or, $-\frac{2v}{v^2 - 1} \, dv = \frac{1}{x} \, dx$ ... (3)

Now, integrating both sides of equation (3), we get,

$$\ln \left( \frac{1}{v^2 - 1} \right) = \ln x + \ln p \quad \text{[where } \ln p \text{ is integration constant]}$$

or, $\ln \left( \frac{x^2}{y^2 - x^2} \right) = \ln (x p)$

or, $y^2 - x^2 = x \left( \frac{1}{p} \right) = c x$ [where, $c = \frac{1}{p} = \text{constant}$]

Answer: Solution is $y^2 - x^2 = c x$

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