Differential Equation (Very peculiar initial condition)

1. Mar 28, 2012

TranscendArcu

1. The problem statement, all variables and given/known data

Solve: $dy/dy = y^3$ given the initial condition y(0)=0

3. The attempt at a solution
$\int \frac{dy}{y^3} = \int dt$
$\frac{-1}{2y^2} = t + c$
$y^2 = \frac{-1}{2(t+c)}$
$y = ± \sqrt{ \frac{1}{2(-t-c)}}$

This equation isn't going to support the initial condition, so can someone tell me where I've gone wrong in calculating the integral?

2. Mar 28, 2012

jackmell

You have:

$$\frac{dy}{dt}=y^3,\quad y(0)=0$$

so immediately I want to divide out by $y^3$. But I can do that only if $y\neq 0$. Poke-a-poke then. However, if $y=0$, then I can't. But if y=0, then what?

3. Mar 28, 2012

TranscendArcu

So if y'=y^3 = 0, then clearly y=0, and it is the constant zero, which makes sense since such a function has y'=0 as well.

4. Mar 28, 2012

jackmell

I think you can say that a little better: if y=0, then $\frac{dy}{dt}=0$ so that $y=k$ is a solution to the DE so that the solution to the IVP:

$$y'=y^3,\quad y(0)=0$$

is y=0.