Question

The solution of dy/dx = $y^2$ with initial value y(0) = 1 bounded in the interval

• A. $-\infty <x<\infty$
• B. $-\infty <x\le 1$
• C. x < 1 , x > 1
• D. $-2\le x\le 2$

Answer 1

The correct option is C x < 1 , x > 1

$\frac{dy}{dx}=y^2$

$\int \frac{dy}{y^2}=\int dx$

$\therefore x=-\frac{1}{y}+c$

$\because y\left(0\right)=1$

$\therefore c=1$

$\therefore y=\frac{1}{1-x}$

But 1 - x $\ne$ 0

x $\ne$ 1

$\therefore xϵ(-\infty ,\infty )-$ {1}

or x > 1

x < 1