​$$\text{Let } y_0 > 0, \ z_0 > 0, \ \alpha > 1. \\[10pt]\text{Consider the following two differential equations:} \\[10pt](*) \quad \frac{dy}{dt} = y^{\alpha} \quad \text{for} \ t > 0, \quad y(0) = y_0 \\[10pt](**) \quad \frac{dz}{dt} = -z^{\alpha} \quad \text{for} \ t > 0, \quad z(0) = z_0 \\[10pt]\text{We say that the solution to a differential equation exists globally if it exists for all } t > 0.$$

Which of the following statements is true?

Correct option is D

​$$\text{Let } y_0 > 0, \ z_0 > 0, \ \alpha > 1. \\[10pt]\text{Consider the following two differential equations:} \\[10pt](*) \quad \frac{dy}{dt} = y^{\alpha} \quad \text{for} \ t > 0, \quad y(0) = y_0 \\[10pt](**) \quad \frac{dz}{dt} = -z^{\alpha} \quad \text{for} \ t > 0, \quad z(0) = z_0 \\[10pt]\text{We say that the solution to a differential equation exists globally if it exists for all } t > 0. \\[10pt]\text{Solution:} \\[10pt]\text{For equation (*)} \\[10pt]\frac{dy}{dt} = y^{\alpha} \\[10pt]\implies \frac{dy}{y^{\alpha}}=t\\[10pt]\text{Integrating both sides:} \\[10pt]\frac{1}{(1 - \alpha)} y^{1-\alpha} = t + c \\[10pt]\text{Using initial condition } y(0)=y_0\\[10pt]\frac{1}{(1 - \alpha)} y_0^{1-\alpha} = c\\[10pt]\frac{1}{(1 - \alpha)} y^{1-\alpha}-\frac{1}{(1 - \alpha)} y_0^{1-\alpha}= t\\[10pt]y^{1-\alpha}-y_0^{\alpha}=(1-\alpha)t\\[10pt]y^{1-\alpha}=t(1- \alpha)-y_0^{1-\alpha}\\[10pt]y^{\alpha-1}=\frac{1}{\frac{1}{y_0^{\alpha-1}}-(\alpha-1)t} \quad \cdots (*)$$

$$\text{For equation (**)}\\[10pt]\frac{dz}{-z^{\alpha}}=dt\\[10pt]\text{Integrating both sides:}\\[10pt]\frac{-z^{(1-\alpha)}}{(1-\alpha)}=t+c\\[10pt]\text{Using initial conditions:}\\[10pt]\frac{-z_0^{(1-\alpha)}}{1-\alpha}=c\\[10pt]\implies \frac{-z^{(1-\alpha)}}{(1-\alpha)}+\frac{z_0^{(1-\alpha)}}{1-\alpha}=t\\[10pt]\text{Simplifing:}\\[10pt]z^{\alpha-1}=\frac{1}{\frac{1}{z_0^{\alpha-1}}+t(\alpha-1)}$$

Now, denominator of (**) never becomes zero so global solution exist,

But denominator of (*) can take value zero so y(t) may tend to infinity as t tends to some T>$$\infty$$

$$\textbf{Hence, Option D is correct.}$$​​​