Question

Solution of diff. equation $(6x + 2y - 10)\frac{dy}{dx}=2x+9y-20$ is

• A. $(y + 2x)^2 = C (x - 2y + 5)$
• B. $(y + 2x)^3 = C (x + 2y - 5)$
• C. $ (y - 2x)^2 = C (x + 2y - 5) $
• D. none of these

Answer 1

The Correct Option is C

$\frac{dy}{dx} = \frac{2x+9y-20}{6x+2y-10}$. Put $x = X + h$ ; $y = Y + k$ $\frac{dY}{dX} = \frac{2X+9Y}{6X+2Y}$ Where $2h + 9k-20 = 0$ $6h + 2k- 10 = 0$ $\frac{h}{-50} = \frac{k}{-100} = \frac{1}{-50}$ $\therefore h = 1$, $k = 2$ Put $Y = vX$ $v+X \frac{dv}{dX} = \frac{2+9v}{6+2v}$ $\therefore X \frac{dv}{dX} = \frac{2+9v-6v-2v^{2}}{6+2v} $ $= \frac{2+3v-2v^{2}}{6+2v}$ $\frac{6+2v}{2+3v-2v^{2}}dv = \frac{dX}{X}$ $\Rightarrow \int \left[\frac{2}{1+2v}+\frac{2}{2-v}\right]dv$ $= \int \frac{dX}{X} + log\,C_{1}$ $\Rightarrow log \left(1 + 2 v\right) - 2 \,log \left(2 - v\right) = log \,C_{1}\, X$ $\Rightarrow log \frac{1+2v}{\left(2-v\right)^{2}} = log\,C_{1}\,X$ $\Rightarrow \frac{1+2 \frac{Y}{X}}{\left(2-\frac{Y}{X}\right) } = C_{1}\,X$ $\Rightarrow \frac{\left(X+2Y\right)X}{\left(2X-Y\right)^{2}} = C_{1}\,X$ $\Rightarrow \left(X + 2Y\right) = \left(2X - Y\right)^{2} C_{1}$ $\Rightarrow C\left(x + 2y - 5\right) - \left(2x - y\right)^{2}$ $\Rightarrow \left(y - 2 x\right)^{2} = C \left(x+ 2y - 5\right)$