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I am trying to remember again the stuff I did about nonlinear differential equations.

I have

$\dot{x}=\left( \begin{matrix}x_1^2 \\ -1\end{matrix}\right)$

I want to solve this nonlinear differential equation and I know that the solution is:

$x_1(t)=\frac{x_1(0)}{1-x_1(0)}$

$x_2(t)=-t+x_2(0)$

I understand how to arrive to the expression of $x_2(t)$ but not to the one of $x_1(t)$.

If I integrate $\dot{x}_1=x_1^2$ I get

$\int\limits_0^t\dot{x}_1(\tau)d\tau=\int\limits_0^t x_1^2(\tau)$

which should give

$x_1(t)-x_1(0)=\left[\frac{x_1^3}{3}\right]^{\tau=t}_{\tau=0}$

which does not give: $x_1(t)=\frac{x_1(0)}{1-x_1(0)}$

Can you help me? Can you give me a link where this procedures are explained? Thanks!

Thanks

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There is independence of equations: $$x_1'=x_1^2,\quad \frac{dx_1}{dt}=x_1^2,\quad\frac{dx_1}{x_1^2}=dt,\quad \int\frac{dx_1}{x_1^2}=\int dt,$$ $$-\frac{1}{x_1}=t+C_1,\quad x_1=\frac{-1}{t+C_1}.$$ On the other hand, $x_2'=-1,\Rightarrow x_2=-t+C_2.$ So, the general solution of the autonomous system is $$\begin{bmatrix}{x_1}\\{x_2}\end{bmatrix}=\begin{bmatrix}{\dfrac{-1}{t+C_1}}\\{-t+C_2}\end{bmatrix}.$$

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  • $\begingroup$ You are welcome. $\endgroup$ – Fernando Revilla Jan 20 '17 at 15:42

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