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Question: The general solution of the differential equation (2x - u + 1)dx +(2y - x + 1)dy = 0 is...

The general solution of the differential equation (2x - u + 1)dx +(2y - x + 1)dy = 0 is

A

x2+y2+xy−x+y=cx^{2} + y^{2} + xy - x + y = c

B

x2+y2−xy+x+y=cx^{2} + y^{2} - xy + x + y = c

C

x2−y2+xy−x+y=cx^{2} - y^{2} + xy - x + y = c

D

x2−y2−2xy+x−y=cx^{2} - y^{2} - 2xy + x - y = c

Answer

x2+y2−xy+x+y=cx^{2} + y^{2} - xy + x + y = c

Explanation

Solution

(2x - y + 1)dx + (2y - x + 1)dx = 0

dydx=2x−y+1x−2y−1,\frac{dy}{dx} = \frac{2x - y + 1}{x - 2y - 1}, put x = X + h, y = Y + k

dYdX=2X−Y+2h−k+1X−2Y+h−2k−1\frac{dY}{dX} = \frac{2X - Y + 2h - k + 1}{X - 2Y + h - 2k - 1}

2h−k+1=0⇒h−2k−1=02h - k + 1 = 0 \Rightarrow h - 2k - 1 = 0

On solving h = -1, k = -1 ∴\therefore dYdX=2X−YX−2Y\frac{dY}{dX} = \frac{2X - Y}{X - 2Y}

Put Y=νX;∴dYdX=ν+XdνdXY = \nu X;\therefore\frac{dY}{dX} = \nu + X\frac{d\nu}{dX}

ν+XdνdX=2X−νXX−2νX=2−ν1−2ν\nu + X\frac{d\nu}{dX} = \frac{2X - \nu X}{X - 2\nu X} = \frac{2 - \nu}{1 - 2\nu}

XdνdX=2−2ν+2ν21−2ν=2(2ν2−ν+1)1−2νX\frac{d\nu}{dX} = \frac{2 - 2\nu + 2\nu^{2}}{1 - 2\nu} = \frac{2(2\nu^{2} - \nu + 1)}{1 - 2\nu}

∴dXX=(1−2ν)2(ν2−ν+1)dν\therefore\frac{dX}{X} = \frac{(1 - 2\nu)}{2(\nu^{2} - \nu + 1)}d\nu

Put ν2−ν+1=t⇒(2ν−1)dν=dt\nu^{2} - \nu + 1 = t \Rightarrow (2\nu - 1)d\nu = dt

∴X=t−1/2c⇒X=(ν2−ν+1)−1/2.c\therefore X = t^{- 1/2}c \Rightarrow X = (\nu^{2} - \nu + 1)^{- 1/2}.c

X2(ν2−ν+1)=constan⁡tX^{2}(\nu^{2} - \nu + 1) = cons\tan t

(x+1)2((y+1)2(x+1)2−(y+1)x+1+1)=constan⁡t(x + 1)^{2}\left( \frac{(y + 1)^{2}}{(x + 1)^{2}} - \frac{(y + 1)}{x + 1} + 1 \right) = cons\tan t

(y+1)2−(y+1)(x+1)+(x+1)2=c(y + 1)^{2} - (y + 1)(x + 1) + (x + 1)^{2} = c

y2+x2−xy+x+y=cy^{2} + x^{2} - xy + x + y = c.