Question

The solutions of $x^2{y}_1^2+xy{y}_1-6y^2=0$ are:

• A. $y=C{x}^2$
• B. $x^{2}y=C$
• C. $\frac{1}{2}\log y=C+\log x$
• D. $x^{3}y=C$

Answer 1

Answer: (A), (C), (D)

$y=C{x}^2$
$\frac{1}{2}\log y=C+\log x$
$x^{3}y=C$

$x^2{y}_1^2+xy{y}_1-6y^2=0$
Solving for $\frac{dy}{dx}$, we get
$(x\frac{dy}{dx}+3y)(x\frac{dy}{dx}-2y)=0$
$\therefore x\frac{dy}{dx}+3y=0\Rightarrow \frac{dy}{y}=-\frac{dx}{x}\Rightarrow y{x}^3=c$
$x\frac{dy}{dx}-2y=0\Rightarrow y=c{x}^2$ or $\frac{1}{2}\log y=c+\log x$