# Help with separable differential equation? $\frac{dy}{dx} =2y^2$

I'm new to separable differential equations, and I'm stuck on this question:$$\dfrac{dy}{dx} = 2y^2$$

Using the initial condition $$y(2)=3$$, find $$y(1)$$.

So far I've integrated to get $$\dfrac{dy}{dx} = \dfrac{2}{3y^3} + C$$. But I'm not sure how to solve for $$C$$ or substitute the initial condition values in, because there's no $$x$$ value given in the original equation. I'm not sure what the next step would be?

• Note: Thinking of $y'$ as $\frac{dy}{dx}$ is helpful in dealing with separation. Apr 4, 2014 at 7:37

$\frac{dy}{dx} = 2y^2$
Move $y^2$ to the opposite side by division. $$\frac{1}{y^2} \cdot \frac{dy}{dx} = 2\\ \int \frac{1}{y^2} dy = \int 2 dx$$
That is not true. $y' = 2y^2$ means that $\dfrac{dy}{dx} = 2y^2$. So $0.5y^{-2}dy = dx$, and integrate both sides with respect to $x$: $\int 0.5y^{-2}dy = \int 1dx$ and this gives:$-0.5y^{-1} = x + C$. So: $y = \dfrac{-0.5}{x + C}$. Now using initial condition $y(2)= 3$ to get: $3 = \dfrac{-0.5}{2 + C}$. So $C = \dfrac{-13}{6}$, and $y = \dfrac{-3}{6x - 13}$, and from this solution we have $y(1) = \dfrac{3}{7}$