# First order nonlinear differential equation 2

The question asks to find y given y(x) is a differentiable function satisfying:

$\frac{dy}{dx}$=-$2xy^4$ , $y(0)$=$\frac{1}{3}$ and $y(x)>0$ , and explain why it is unique.

I assume the steps are to integrate $\frac{dy}{dx}$ to get a function for y(x), then use $y(0)=1/3$ to find the constant and then set $y(x)>0$ to find y but i keep getting stuck.

A differential equation solution gives $y(x)$=$\frac{1}{\sqrt[3]{c_1+3x^2}}$ , which I can use $y(0)$ to find $c_1$ but then I'm not sure how to get y from that since the equation doesn't have any y's in it.

A similar question I found on this site showed to rearrange and integrate each side like $\int$$\frac{dy}{y^4}=\int$$-2xdx$ which gives $-\frac{1}{3y^3}+c_1$=$-x^2+c_2$ , but then I don't know how to get $y(x)$ from that frunction, it seems further from the solution than my first try.

That's just a simple manipulation: \begin{equation*} \begin{split} -\frac{1}{3y^3}+c_1=-x^2+c_2 & \iff -\frac{1}{3y^3} = -x^2 + c_2-c_1 \\ & \iff \frac{1}{3y^3} = x^2 +c_1-c_2 \\ & \iff y^3 = \frac{1}{3x^2 + 3(c_1-c_2)} \\ & \iff y = \frac{1}{\sqrt[3]{3x^2+3(c_1-c_2)}}. \end{split} \end{equation*} Now it's just note that $3(c_1-c_2)$ is constant, say $k$, and we obtain $$y=\frac{1}{\sqrt[3]{3x^2+k}}.$$
• What did you mean by "(...) set $y(x) > 0$ to find $y$ (...)" ? As I see it, $y$ is just a function, not an incognite. – Rodrigo Dias Oct 5 '17 at 23:01