- #1

TranscendArcu

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## Homework Statement

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

## The Attempt at a Solution

[itex]\int \frac{dy}{y^3} = \int dt[/itex]

[itex]\frac{-1}{2y^2} = t + c[/itex]

[itex]y^2 = \frac{-1}{2(t+c)}[/itex]

[itex]y = ± \sqrt{ \frac{1}{2(-t-c)}}[/itex]

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