Solve the equations $$\dfrac{dy}{dx} = \dfrac{y^2-x^2}{2xy}$$

Asked by Sakshi | 1 year ago |  61

##### Solution :-

From the question it is given that,

$$\dfrac{dy}{dx}=\dfrac{ (y^2 – x^2)}{2xy}$$

The given differential equation is a homogeneous equation,

Let us assume, y = vx and $$(\dfrac{dy}{dx}) = v + x (\dfrac{dv}{dx})$$

$$v + x (\dfrac{dv}{dx})= \dfrac{(v^2x^2 – x^2)}{(2xvx)}$$

Then, $$x=\dfrac{dv}{dx}=\dfrac{ (v^2 – 1)}{2v – \dfrac{v}{1}}$$

$$x(\dfrac{dv}{dx}) =\dfrac{ (-1 – v^2)}{2v}$$

Now, taking like variables on same side,

$$\dfrac{2v}{v^2 + 1} dv =\dfrac{ – dx}{x}$$

Integrating on both side we get,

$$∫\dfrac{2v}{v^2 + 1}dv = – ∫\dfrac{dx}{x}$$

log (1 + v2) = – log x + log c

$$1 + v^2 = \dfrac{c}{x}$$

Now substitute the value of v,

$$1 + \dfrac{y^2}{x^2} = \dfrac{c}{x}$$

x2 + y2 = cx

Answered by Aaryan | 1 year ago

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