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#1
Sep 10 2014 02:18am
Solve the given initial-value problems.

1. dy/dt = (y^3) , y(0) = 0

2. dy/dt = t((y^2) + 4), y(0) = 2

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#2
Sep 10 2014 09:05am
can help for 20 fg
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#3
Sep 10 2014 09:21am
Quote (JDota72 @ 10 Sep 2014 17:05)
can help for 20 fg


Its against the rules asking for fg I believe

Anyway here it is for free (no guarantees its correct but it should be)

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#4
Sep 10 2014 09:24am
nm

This post was edited by JDota72 on Sep 10 2014 09:24am
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#5
Sep 10 2014 11:46am
Given your initial conditions, a solution remains for the first : y(t) = 0.
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