1. Mar 24, 2012

### jwqwerty

1. The problem statement, all variables and given/known data
solve dy/dt+y^3=0 , y(0)=y0 and determine how the interval in which the solution exists depends on the initial value y0

2. Relevant equations
Theorem: Let the fuctions f and df/dy(partial differentiation) be continuous in some rectangle a<t<b , c<y<d containing (t0,y0). Then in some interval t0-h<t<t0+h contained in a<t<b, there is a unique solution y=∅(t) of the initial value problem dy/dt=f(t,y), y(t0)=y0

3. The attempt at a solution
According to the theorem above, by solving dy/dt+y^3=0 y should only have one solution (y) in some interval.

dy/dt+y^3=0

i) when y≠0
-y^(-3)dy=dt and integrate both sides
0.5y^(-2)=t+c (c is a constant)

by initial condition y(0)=y0
0.5y^(-2)=t+0.5y0^(-2) and multiply both sides by 2
y^(-2)=2t+y0^(-2)
y^2=1/(2t+y0^(-2))
=y0^2/(2ty0^(2)+1)
Thus, y= ±y0/√(2ty0^(2)+1) (∞>t>-1/2y0^2 since numbers inside root must be bigger than 0)

ii) y=0
y=0 holds true for all t.