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Question

Question: If y¢ = \(\frac{x - y}{x + y}\), then its solution is –...

If y¢ = xyx+y\frac{x - y}{x + y}, then its solution is –

A

y2 + 2xy – x2 = c

B

y2 + 2xy + x2 = c

C

y2 – 2xy – x2 = c

D

y2 – 2xy + x2 = c

Answer

y2 + 2xy – x2 = c

Explanation

Solution

Given, dydx=xyx+y\frac{dy}{dx} = \frac{x - y}{x + y} This is a homogeneous equation

Put y = vx Ž dydx\frac{dy}{dx} = v + x dvdx\frac{dv}{dx}

Given equation becomes

v + x dvdx\frac{dv}{dx} = 1v1+v\frac{1 - v}{1 + v}

Ž xdvdx\frac{dv}{dx} = 1v1+v\frac{1 - v}{1 + v} – v

Ž 1+v2(1+v)2\frac{1 + v}{2 - (1 + v)^{2}} dv = dxx\frac{dx}{x}

On integrating both sides

1+v2(1+v)2\int_{}^{}\frac{1 + v}{2 - (1 + v)^{2}} dv = dxx\int_{}^{}\frac{dx}{x}

Put (1 + v)2 = t Ž 2(1 + v) dv = dt

Ž 12\frac{1}{2} dt2t\int_{}^{}\frac{dt}{2 - t} = dxx\int_{}^{}\frac{dx}{x}

Ž –12\frac{1}{2}log (2 – t) = log x + log c

Ž –12\frac{1}{2}log [2 – (1 + v)2] = log xc

Ž –12\frac{1}{2} log [–v2 – 2v + 1] = log xc

Ž log 112vv2\frac{1}{\sqrt{1 - 2v - v^{2}}} = log xc

Ž x2c2 (1 – 2v – v2) = 1

Ž x2c2 (12yxy2x2)\left( 1 - \frac{2y}{x} - \frac{y^{2}}{x^{2}} \right) = 1[v=yx]\left\lbrack \because v = \frac{y}{x} \right\rbrack

Ž x2c2(x22yxy2)x2\frac{x^{2}c^{2}(x^{2} - 2yx - y^{2})}{x^{2}} = 1

Ž y2 + 2xy – x2 = c