Question

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

• A. $y=C{x}^2$
• B. $x^{3}y=C$
• C. $\frac{1}{2}\ln y=C+\ell nx$
• D. All of these

Answer 1

Answer: (D)

$x^2{y}_1^2+xy{y}_1-6y^2=0$
It is quadratic equation in $y_1$
$y_1=\frac{-xy\pm \sqrt{x^2{y}^2+24y^2{x}^2}}{2x^2}=\frac{-xy\pm 5xy}{2x^2}$
$y_1=-\frac{3y}{x}\mid y_1=\frac{2y}{x}$
$\frac{dy}{dx}=\frac{-3y}{x}\mid \frac{dy}{dx}=\frac{2y}{x}$
$-\frac{dy}{y}=3\frac{dx}{x}|\frac{dy}{dx}=\frac{2y}{x}$
$-\ell \ln y=3\ln x+\ln c|\ln y=2\ln x+\ell nc$
$x^{3}y=Cy=c{x}^2$
Option (3)
$\ln y=c+2\ell \ln x$
$\frac{1}{2}\ln y=\ell n{c}_1+\ell nx$