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(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)

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    $\begingroup$ For part (b) show that your solution from part (a) blows up in finite time. $\endgroup$ – Jacky Chong Sep 26 '16 at 4:26
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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 $$I=\left(0, \frac{x_0^{1-n}}{n-1}\right).$$

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  • $\begingroup$ @Math1000..thanks $\endgroup$ – RAJ Sep 26 '16 at 6:04
  • $\begingroup$ how to verify that $x\in I$ approaches the finite endpoint of I $x(t)->\infty$ and that $t\in I$ approaches the non fine endpoint of I, x(t)->0 $\endgroup$ – RAJ Sep 26 '16 at 6:06
  • $\begingroup$ @math1000..can you help me for the above problem too $\endgroup$ – RAJ Sep 26 '16 at 6:59

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