# 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. – Display Name Apr 4 '14 at 7:37

## 2 Answers

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

Hint:

Move $y^2$ to the opposite side by division. $$\frac{1}{y^2} \cdot \frac{dy}{dx} = 2\\ \int \frac{1}{y^2} dy = \int 2 dx$$

Note: Even though here you can see that dx is 'multiplied' to the otherside(which is the easiest way to think of it), this does not hold in general, but will never prove to be a problem in regards to separation.

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 $y(1) = \dfrac{3}{7}$