Question

If dy/dx equals xy squared and if y = 1 when x = 0, then when y = 3, x is equal to?

Answer 1

$\displaystyle \dfrac{dy}{dx}=xy^2\\ dy=xy^2 \, dx\\ \dfrac{1}{y^2}\, dy=x\, dx\\ \int \dfrac{1}{y^2}\, dy=\int x\, dx\\ -\dfrac{1}{y}=\dfrac{x^2}{2}+C\\\\ -\dfrac{1}{1}=\dfrac{0^2}{2}+C\\ C=-1\\\\-\dfrac{1}{3}=\dfrac{x^2}{2}-1\\ -2=3x^2-6\\ 3x^2=4\\ x^2=\dfrac{4}{3}\\ x=\sqrt{\dfrac{4}{3}} \vee x=-\sqrt{\dfrac{4}{3}}\\ x=\dfrac{2}{\sqrt3} \vee x=-\dfrac{2}{\sqrt3}\\ x=\dfrac{2\sqrt3}{3} \vee x=-\dfrac{2\sqrt3}{3}$

Answer 2

This is a separable differential equation, so let's start of there. Let's separate the variables to their own side with the respective differentials:
$\frac{dy}{dx} = xy^2$
$dy = (xy^2) dx$
$\frac{1}{y^2} dy = x dx$

Let's integrate both sides (it's separable, so we can do this):
$\int\ { \frac{1}{y^2} } \, dy = \int\ {x} \, dx$
$- \frac{1}{y} = \frac{x^2}{2} + C$

Now, let's plug in the values we are given to find the constant "C":
$- \frac{1}{1} =\frac{0^2}{2}+C$
$-1 = C$

Let's rewrite the equation, with C in it, then solve for x because we need to ultimately find x:
$- \frac{1}{y} = \frac{x^2}{2} - 1$
$x = \sqrt{2(- \frac{1}{y}+1)}$

Let's plug in y = 3 and solve for x:
$x = \sqrt{2(- \frac{1}{3}+1)} = \sqrt{ 2( \frac{2}{3}) } = \sqrt{ \frac{4}{3} }$

Let's simplify and rationalize the denominator:
$x = \sqrt{ \frac{4}{3}} = 2 \sqrt{ \frac{1}{3}} = 2 \frac{ \sqrt{3} }{3}$

So, your answer is D.