(a) Let $n\geq2$ be an integer and $x_o$ a areal number.obtain explicit the initial solution of the initial value problem $\mathring{x}=x^n ,\;x(0)=x_0$
(b) In part (a) assume that $x_o>0$, verify answer to part(a) that the solutions of the initial value problem . $\mathring{x}=x^n,\;(0)=x_o$ exists an proper interval I containing '0' and that I is a proper subset of $\mathbb{R}(I\neq \mathbb{R})$
my attempt: $\frac{dx}{dt}=x^n\\ \frac{dx}{x^n}=dt\\ \frac{x^{-{n}+1}}{-n+1}=t+c\\ \text{as initival value condition};\frac{x^{-{n}+1}}{-n+1}=\frac{x(0)^{-{n}+1}}{-n+1}+t $
is ia m right then how to solve part (b)
Continuing your computations, I found that $$x(t) = \left(x_0^{1-n} + (1-n)t \right)^{\frac1{1-n}}. $$ We have $$x(0) = \left(x_0^{1-n}\right)^{\frac1{1-n}}=x_0>0,$$ and since $\lim_{y\to0+}y^{\frac1{1-n}}=+\infty,$ we find from $$x_0^{1-n} + (1-n)t >0\iff t<\frac{x_0^{1-n}}{n-1} $$ that the solution $x$ exists on the open interval $$I=\left(0, \frac{x_0^{1-n}}{n-1}\right).$$