(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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?

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# Homework Help: About differential question

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