Question

Solution of the equation $x^2 dy + y (x + y) dx = 0$ is

• A. $ y + 2x = C^2 x^2 y $
• B. $y-2x=\frac{C^2x^2}{y}$
• C. $y+2x=\frac{C^2x^2}{y}$
• D. none of these.

Answer 1

The Correct Option is A

$x^{2}\,dy+y\left(x + y\right)dx = 0$ $\Rightarrow \frac{dy}{dx} = -\frac{y\left(x+y\right)}{x^{2}}$, homogeneous. Put $y = vx$ $\therefore\frac{dy}{dx}=v+x \frac{dv}{dx}$ $\therefore v+x \frac{dv}{dx} = -\frac{vx\left(x+vx\right)}{x^{2}} = -v-v^{2}$ $\therefore x \frac{dv}{dx} = -2\,v-v^{2}$ $\therefore \frac{dv}{-v\left(2+v\right)} = \frac{dx}{x}$ $\Rightarrow \frac{dv}{v\left(v+2\right)}+\frac{dx}{x}=0$ $\Rightarrow \frac{1}{2} \int \left(\frac{1}{v}-\frac{1}{v+2}\right)dv+\int \frac{dx}{x} = log\,C$ $\Rightarrow \frac{1}{2} log \frac{v}{v+2} + log\,x=log\,C$ $\Rightarrow log \frac{v}{v+2} + log\,x^{2}=log\,C$ $\Rightarrow \frac{v}{v+2}.x^{2} = C$ $\Rightarrow \frac{\frac{y}{x}}{\frac{y}{x}+2}.x^{2} = C$ $\Rightarrow \frac{y}{y+2x}.x^{2}=C$ $\Rightarrow y+2x = \frac{1}{C}x^{2}y$ $= C^{2}x^{2}y$ (Take $\frac{1}{C}= C^{2}$) $\Rightarrow y + 2x=C^{2}x^{2}y$.