Question

The solution of the differential equation $\frac{dy}{dx}=-\left(\frac{x-2y+5}{2x-y+4}\right)$ is
(where $C$ is integration constant)

• A. $(x+y-1{)}^3=C(y-x-3)$
• B. $(x+y+3)=C(x+y-1{)}^3$
• C. $(x+y-3)=C(x-y+1{)}^3$
• D. ${(x+y-1)}^3=C(x+y-3)$

Answer 1

The correct option is A $(x+y-1{)}^3=C(y-x-3)$

Put $x=X+h$ and $y=Y+k$ in
$\frac{dy}{dx}=-\left(\frac{x-2y+5}{2x-y+4}\right)$
Such that
$h-2k+5=0$ &
$2h-k+4=0$
On solving, we have $h=-1,k=2$
$\Rightarrow \frac{dy}{dx}=\frac{dY}{dX}=-\frac{X-2Y}{(2X-Y)}$
Put $Y=vX$
$\Rightarrow \frac{dY}{dX}=v+\frac{Xdv}{dX}=\frac{1-2v}{-(2-v)}$
$\Rightarrow \frac{Xdv}{dX}=\frac{1-2v}{v-2}-v$
$\Rightarrow \frac{Xdv}{dX}=\frac{-(v^2-1)}{v-2}$
$\Rightarrow \int \frac{v-2}{(v^2-1)}dv=-\int \frac{dX}{X}$
$\Rightarrow \int [\frac{-1}{2(v-1)}+\frac{3}{2(v+1)}]dv=-\int \frac{dX}{X}$
$\Rightarrow -\frac{1}{2}\ln |v-1|+\frac{3}{2}\ln |v+1|=-\ln |X|+\ln |c|$
$\Rightarrow \ln \left|\frac{(v+1{)}^{3/2}}{(v-1{)}^{1/2}}\right|=\ln \left|\frac{c}{X}\right|$
$\Rightarrow \ln \left|\frac{(Y+X{)}^{3/2}}{X(Y-X{)}^{1/2}}\right|=\ln \left|\frac{c}{X}\right|$
$\Rightarrow |(Y+X{)}^{3/2}|=|c(Y-X{)}^{1/2}|$
$\Rightarrow (Y+X{)}^3=C(Y-X\left)\right(\text{let}{c}^2=C)$
$\Rightarrow (x+y-1{)}^3=C(y-x-3)$