Let y₀ > 0, z₀ > 0 and α > 1.

Consider the following two differential equations:
\(\begin{aligned} &(*)\left\{\begin{array}{l} \frac{d y}{d t}=y^\alpha \quad \text { for } t>0, \\ y(0)=y_0 \end{array}\right. \\ &(* *)\left\{\begin{array}{l} \frac{d z}{d t}=-z^\alpha \quad \text { for } t>0, \\ z(0)=z_0 \end{array}\right. \end{aligned}\)
We say that the solution to a differential equation exists globally if it exists for all t > 0.

Which of the following statements is true?

Explanation:

 \(\frac{d y}{d t}=y^α\)

\(\Rightarrow \frac{y^{1-α}}{1-α}=t+c\)

and y(0) = y₀ ⇒ \(\frac{y_0^{1-α}}{1-α}=c\)

then y1-α = (1 - α)t + y01-α

and If α > 1 ⇒ 1 - α < 0. suppose 1 - α = -a, a > 0

then (1) y-a = -at + y0-a 

⇒ \(y^a=\frac{1}{y_0^{-a}-a t}\)

then for y0-a - at = 0, solution does not exist.

\(\Rightarrow \quad t=\frac{y_o^{-a}}{a}=\frac{1}{a \cdot y_o^a}>o\)

(∵ y0 > 0, a > 0) 

∴ (*) has no global solution

as \(\frac{1}{a \cdot y_0^a}\) ∈ (0, ∞)

options (1) and (3) are false

(* *) \(\frac{d z}{d t}=-z^α\) ⇒ \(\frac{z^{1-α}}{1-α}=-t+c\)

and z0 = z(0) ⇒ \(\frac{z_0^{1-α}}{1-α}=c\)

∴ z1-α = -(1 - α)t + z01-α ....(ii)

and for α > 1 ⇒ 1 - α < 0. So, Suppose 1 - α = - b, b > 0

then (ii) ⇒ z-b = bt + zo-b ⇒ \(z^b=\frac{1}{b t+z_0^{-b}}\)

and for bt + z0-b​ = 0 solution does not exist 

⇒ \( t=-\frac{z_0^{-b}}{b}=-\frac{1}{z_0^b \cdot b}\) < o

(∵ zo > o, b > 0)

So, ∀ t > 0, solution exist of (* *)

⇒​ (**) have global solutions.

option (2) is false.

Also, for T < ∞, take T = \(\frac{1}{a \cdot y_0 a}\)

\(\lim _{t \rightarrow T}|y(t)|=\lim _{t \rightarrow T}\left|\left(\frac{1}{y_0^{-a}-a t}\right)^{1 / a}\right| \)

\(=\lim _{t \rightarrow \frac{1}{a \cdot y_0} a}\left|\left(y_0^{\frac{1}{-a}-a t}\right)^{1 / a}\right|\)

= + ∞ 

∴ option (4) is true.