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 $Y = v\, X$
$\therefore v + \frac{Xdv}{dX}+\frac{3+2v}{2+3v} = 0$
$\therefore X \frac{dv}{dX}+\frac{3+2v+2v+3v^{2}}{2+3v} = 0$
$\therefore \frac{3v+2}{3v^{2}+4v+3}dv + \frac{dX}{X} = 0$
$\Rightarrow \frac{6v+4}{3v^{2}+4v+3} dv + \frac{2dX}{X} = 0$
$\therefore log \left(3v^{2} + 4v + 3\right) + 2\, log \, X = log \, C$
$\Rightarrow log \left(3 \frac{Y^{2}}{X^{2}}+4 \frac{Y}{X}+3\right)+log\, X^{2}= log\, C$
$\Rightarrow log \left(3Y^{2} + 4YX + 3X^{2}\right) = log \,C$
$\Rightarrow 3Y^{2} + 4YX + 3X^{2} = C$
$\Rightarrow 3\left(y - 1\right)^{2} + 4 \left(y - 1\right) \left(x - 1\right) + 3 \left(x- 1\right)^{2} = C_{1}$
$\Rightarrow 30\left(x^{2} + y^{2}\right) + 4xy - 10 \left(x + y\right) = k$
Where $k = C_{1} - 7$