Question: Determine all differentiable functions in the form $y$ = $f(x)$ which have the properties:

$f'(x)$=$(f(x))^3$ and $f(0)=2$

What I have done

I set up the differential equation as such

$$ \frac{dy}{dx} = y^3 $$

$$ \int y^{-3} dy = \int dx$$

$$ {-1\over 2y^2} = x + c $$

$$ {1\over 2y^2} = -x-c $$

At $y = 2 , x = 0 , c = -\frac{1}{8}$

$$ {1\over 2y^2} = {1\over 8} - x$$

$$ 2y^2 = {8\over {1-8x}} $$

$$ y^2 = {4\over {1-8x}} $$

$$ y = ±\sqrt{4\over {1-8x}} $$

Is this correct? Are there any limits of x or y to consider?

  • $\begingroup$ Yes it is okay. $\endgroup$ – SchrodingersCat Dec 28 '15 at 6:15
  • 2
    $\begingroup$ But if the sign is negative then is your initial condition true? $\endgroup$ – Elliot G Dec 28 '15 at 6:21
  • $\begingroup$ @ElliotG Therefore the conditions is y = the positive equation not negative and x =/= 1/8? $\endgroup$ – bigfocalchord Dec 28 '15 at 6:35
  • $\begingroup$ @dydxx That is correct $\endgroup$ – GaussTheBauss Dec 28 '15 at 6:46

This is a very good solution and I detect no flaws in it. However, by taking the square root, you have introduced a superfluous solution ($y = -\sqrt{4\over {1-8x}}$) into the set of solutions. The solution here is unique.

I don't know the source but let me paraphrase a brief conjecture relevant here.

An initial value problem for a first order differential equation either has one function as a solution or no solutions.

I won't go into when there are not any solutions. That's a pretty complicated discussion regarding intervals and things. However, at the core of this question you must recognize there is one solution and that solution is:

$$y = f(x) = \sqrt{4\over {1-8x}}$$


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