Differential Equation (Very peculiar initial condition)

[TranscendArcu]

Homework Statement

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

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?

 

[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?

 

[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.

 

[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.