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 $$y=\frac{1}{\sqrt[3]{3x^2+k}}.$$

  • $\begingroup$ Is that all I need as the complete answer? No need to use the given properties/conditions? $\endgroup$ – S. Flemming Oct 5 '17 at 22:53
  • $\begingroup$ The given differential equation is separable, and the method to solve separable first order differential equations is as you've mentioned in your question. What I have done is manipulate your partial solution to get the solution you said don't know how to get it. Am I missing something? $\endgroup$ – Rodrigo Dias Oct 5 '17 at 22:57
  • $\begingroup$ What did you mean by "(...) set $y(x) > 0$ to find $y$ (...)" ? As I see it, $y$ is just a function, not an incognite. $\endgroup$ – Rodrigo Dias Oct 5 '17 at 23:01
  • $\begingroup$ Judging by the conditions the question gave I assumed I would end up with an equation y(x) in terms of y and x with a constant that I would need to compute using y(0)=1/3, and then set that equation <0 to get y<some function of x. $\endgroup$ – S. Flemming Oct 5 '17 at 23:01
  • $\begingroup$ I'm thinking that y(x) and y are two different things, are they not? $\endgroup$ – S. Flemming Oct 5 '17 at 23:06

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