How Does the Initial Value y0 Affect the Solution Interval of dy/dt + y^3 = 0?

[jwqwerty]

Homework Statement

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

Homework 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

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.

Then there is a contradiction
1. y has three solutions when ∞>t>-1/2y0^2 which are y= ±y0/√(2ty0^(2)+1), y=0.

did i do something wrong?

 

[mighty2000]

No, your solution is correct. The initial value y0 affects the solution in two ways:
1. If y0=0, then y=0 for all values of t.
2. If y0≠0, then y has two solutions, y=±y0/√(2ty0^(2)+1), in the interval ∞>t>-1/2y0^2.