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# General solution of differential equation: $\large\frac{dy}{dx}=\frac{y+x-2}{y-x-4}$

$(a)\;x^2+2xy-y^2-4x+8y=c \\ (b)\;x^2+2xy+y^2+4x+8y=c \\ (c)\;x^2+2xy-y^2+4x+8y=c \\ (d)\;x^2+2xy+y^2-4x+8y=c$
Can you answer this question?

$\large\frac{dy}{dx}= \frac{y+x-2}{y-x-4}$
$x= X+n\qquad y= Y+k$
$dx=dX \qquad dy = dY$
$\large\frac{dY}{dX}= \frac{X+h+Y+k-2}{Y+k-X-h-4}$
$k+h-2=0$
$k-h-4=0$
$k=3 \qquad h=-1$
$\large\frac{dY}{dX}=\large\frac{X+Y}{X-Y}$
$\large\frac{Y}{X}$$=u Y=uX \large\frac{dY}{dX}$$=u+ X \large\frac{du}{dX}$
$u+ X \large\frac{dy}{dX}=\large\frac{u+1}{u-1}$
$X \large\frac{dy}{dX}= \frac{u+1}{u-1}-u$
$\qquad= \large\frac{u-1-u^2+u}{u-1}$
$\int - X \large\frac{dy}{dx} =\int \large\frac{u^2-2u-1}{u-1}$
$\int - X \large\frac{(u-1)}{u^2-2u-1}$$du =\int \large\frac{dx}{x} \int - \large\frac{(2u-2)du}{u^2-2u-1} =\int \large\frac{2dx}{x} u^2-2u-1=t (2u-2)=\large\frac{dt}{du} - \int \large\frac{dt}{t}$$ =2 \int \large\frac{dx}{x}$
$-\log t = 2\log x+c$
$2 \log x+ \log t=c$
$\log (Y^2-2XY-X^2)=c$
$Y^2+2XY-T^2=e^c=-c'$
$X^2+2XY-Y^2=c'$
Replacing $X=x+1;Y=y-3$
$(x+1)+2 (x+1)(y-3)-(y-3)^2=c'$
$x^2+2xy-y^2-4x+8y-14=c'$
$x^2+2xy-y^2-4x+8y=c$
Hence a is the correct answer.
answered Feb 4, 2014 by