 doodverft05

2022-06-20

I am trying to remember again the stuff I did about nonlinear differential equations.
I have
$\stackrel{˙}{x}=\left(\begin{array}{c}{x}_{1}^{2}\\ -1\end{array}\right)$ Kaydence Washington

I am trying to remember again the stuff I did about nonlinear differential equations.
I have
$\stackrel{˙}{x}=\left(\begin{array}{c}{x}_{1}^{2}\\ -1\end{array}\right)$
I want to solve this nonlinear differential equation and I know that the solution is:
${x}_{1}\left(t\right)=\frac{{x}_{1}\left(0\right)}{1-{x}_{1}\left(0\right)}$
${x}_{2}\left(t\right)=-t+{x}_{2}\left(0\right)$
I understand how to arrive to the expression of ${x}_{2}\left(t\right)$ but not to the one of ${x}_{1}\left(t\right)$.
If I integrate ${\stackrel{˙}{x}}_{1}={x}_{1}^{2}$ I get
$\underset{0}{\overset{t}{\int }}{\stackrel{˙}{x}}_{1}\left(\tau \right)d\tau =\underset{0}{\overset{t}{\int }}{x}_{1}^{2}\left(\tau \right)$
which should give
${x}_{1}\left(t\right)-{x}_{1}\left(0\right)={\left[\frac{{x}_{1}^{3}}{3}\right]}_{\tau =0}^{\tau =t}$
which does not give: ${x}_{1}\left(t\right)=\frac{{x}_{1}\left(0\right)}{1-{x}_{1}\left(0\right)}$
Can you help me?

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