Question

Solution of the differential equation $\frac{dy}{dx}+\frac{3x+2y-5}{2x+3y-5}=0$ is

• A. $3x^2+4xy+3y^2-10x-10y=k$
• B. $x^2+4xy-y^2-4x+6y=k$
• C. $(x+2y{)}^2+3y=k$
• D. none of these.

Answer 1

Answer: (A)

Put $x=X+h$, $y=Y+k$
$\therefore dx=dX$, $dy=dY$
$\Rightarrow \frac{dy}{dx}=\frac{dY}{dX}$
$\frac{dY}{dX}+\frac{3X+2Y+3h+2k-5}{2X+3Y+2h+3k-5}=0$
Put $3h+2k-5=0$
$2h+3k-5=0$
$\therefore \frac{h}{5}=\frac{k}{5}=\frac{1}{5}$
$\therefore h=1$, $k=1$
$\therefore \frac{dY}{dX}+\frac{3X+2Y}{2X+3Y}=0$
Put $Y=vX$
$\therefore v+\frac{Xdv}{dX}+\frac{3+2v}{2+3v}=0$
$\therefore X\frac{dv}{dX}+\frac{3+2v+2v+3v^2}{2+3v}=0$
$\therefore \frac{3v+2}{3v^2+4v+3}dv+\frac{dX}{X}=0$
$\Rightarrow \frac{6v+4}{3v^2+4v+3}dv+\frac{2dX}{X}=0$
$\therefore log\left(3v^2+4v+3\right)+2logX=logC$
$\Rightarrow log\left(3\frac{Y^2}{X^2}+4\frac{Y}{X}+3\right)+log{X}^2=logC$
$\Rightarrow log\left(3Y^2+4YX+3X^2\right)=logC$
$\Rightarrow 3Y^2+4YX+3X^2=C$
$\Rightarrow 3{\left(y-1\right)}^2+4\left(y-1\right)\left(x-1\right)+3{\left(x-1\right)}^2=C_1$
$\Rightarrow 30\left(x^2+y^2\right)+4xy-10\left(x+y\right)=k$
Where $k=C_1-7$