Interval of validity of a non-linear differential equation

 In the following problem we are going to determine the interval of validity of a non-linear differential equation.

Problem. Determine the interval of validity of the initial value problem and give its dependence on the value of y₀.

y' = y²  y(0) = y₀

Let's see if f(t, y) = y² and δf/δy are continuous. It's obvious that f(t, y) =  y² is continuous. 

Let's calculate δf/δy:

δf/δy = 2y. It's continuous. Then there is a unique solution to the IVP in an interval that contains t₀ = 0

We have to determine this solution in order to find the interval of validity. 

Let's solve the differential equation

dy/dt = :y² 

dy = y² dt

dy/ y² = dt

y^-2 dy = dt

∫y^-2 dy = ∫dt

y^-2+1/-2 +1 = t + c

y^-1/-1 = t

-1/y = t + c

y = -1/t + c

Let's determine the value of c. Let's substitute t by 0 and y by  y₀ as it's given in the initial condition

y₀ = -1/c c = -1/y₀

Let's substitute c in y we have:

y = -y₀ /y₀ t -1

The interval of validity of the IVP is the interval where y is defined. 

Let's determine this interval :of validity.

The function y isn't defined if the denominator is equal to 0.

If y₀ t -1 = 0, t = 1/y₀ . Therefore y isn't defined for that value of t.

Y is defined in the following intervals: -∞<t<1/y₀  and 1//y₀<t<+∞

Let's determine the interval of validity for different values of y₀:

If y₀ = 0, then the interval of validity is -∞<t<+∞

If y₀<0  1/y₀<0 t = 0 doesn't belong to  -∞<t<1/y₀ but belongs to 1//y₀<t<+∞, which is an interval of validity

If y₀>0 then 1/y₀>0  -∞<t<1/y₀

The problem that we just solved demonstrates the difference between the interval of validity of linear differential equations and that of non-linear differential equations. The interval of validity of linear differential equations doesn't depend the value of y₀. The interval of validity of non-linear differential equations depends on the value of  y₀.