Question

The particular solution of differential equation $\frac{dy}{dx}=-4x{y}^2,y\left(0\right)=1$ is ______

• A. $y=(2x^2+1)=1$
• B. $x^2=\frac{1}{y^2}$
• C. $y=x^2+\log x$
• D. $4e^x+\frac{1}{y^2}=8$

Answer 1

Answer: (A)

$y=(2x^2+1)=1$

$y={(2x^2+1)}^{-1}$
$\frac{dy}{dx}=-4x{y}^2$
$\therefore \frac{dy}{y^2}=-4xdx$
$\therefore \int y^{-2}dy=-4\int xdx+c$
$\therefore \frac{y^{-1}}{-1}=-4\left(\frac{x^2}{2}\right)+c$
$\therefore \frac{-1}{y}=-2x^2+c$
Take $x=0$ and $y=1$
$\therefore -1=-2\left(0\right)+c$
$\therefore c=-1$
$\therefore \frac{-1}{y}=-2x^2-1$
$\therefore \frac{1}{y}=1+2x^2$
$\therefore y={(1+2x^2)}^{-1}$