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  1. Show that the differential equation $x^.=x^1/5$ with initial condition $x(0)=0$ has non unique solutions. Why does the theorem on uniqueness of solutions not apply?

$$x(t)=(\frac{4(t-t_0)}{5})^{5/4}$$

$=z(t;t_0)=0$ for $t\le\ t_0$ $=z(t;t_0)=(\frac{4(t-t_0)}{5})^{5/4}$ for $t\ge\ t_0$

The theorem on uniqueness of solutions does not apply because $f'(0)$ is not defined

  1. Consider the differential equation $x^.=x^2$ (sorry I don't know how to format the x dot), with initial condition $x(0)=x_0≠0$.

a. Verify that the theorem on existence and uniqueness applies.

For this step, I am not really sure what to do? For a previous question there was an example that the theorem on existence and uniqueness fails because $f'(0)$ was undefined, so for this part I just wrote down:

$x^.=x^2$ $f(x)=x^2$ $f'(x)=2x$ $f'(0)=2(0)=0$

My book didn't really do a good job of explaining this theorem, so I am not really sure what the requirements are.

b. Solve for an explicit solution. $$x^.=x^2$$ $$f(x)=x^2$$ $$F(x)=∫1/f(x)=∫dt$$ $$=∫1/x^2$=∫dt$$ $$=∫x^{-2}$=∫dt$$ $$=-x^-1=t-t_0$$ $$=x^-1=t_0+t$$ $$=x(t)=\frac{1}{t_0+t}$$

c. What is the maximal interval of definition if $x_0>0$?

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    $\begingroup$ You can find 2. solved here. $\endgroup$ – Git Gud Oct 31 '14 at 17:30
  • $\begingroup$ $\LaTeX$ for $\dot x$ is "\dot x"; and don´t forget your dollar signs! Cheers' $\endgroup$ – Robert Lewis Oct 31 '14 at 17:37

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