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Given nonlinear first order ordinary differential equation \begin{eqnarray} \dfrac{dx_1}{dt}&=&4x_1\left(1-\dfrac{x_2}{2}\right)\\ \dfrac{dx_2}{dt}&=&3x_1\left(\dfrac{x_2}{3}-1\right)\\ \end{eqnarray} with initial value $x_1(0)=3$ and $x_2(0)=5$, $0\leq t\leq 1$.

I can't solve that because the system is nonlinear. I only studied to solve linear system of ODE.

How to solve that nonlinear first order ordinary differential equation?

\begin{eqnarray} \dfrac{dx_1}{dt}&=&4x_1\left(1-\dfrac{x_2}{2}\right)\\ \dfrac{dx_2}{dt}&=&3x_1\left(\dfrac{x_2}{3}-1\right)\\ \end{eqnarray} \begin{eqnarray} \dfrac{\dfrac{dx_1}{dt}}{\dfrac{dx_2}{dt}}=\dfrac{4x_1\left(1-\dfrac{x_2}{2}\right)}{3x_1\left(\dfrac{x_2}{3}-1\right)}\\ \dfrac{dx_1}{dx_2}=\dfrac{4\left(\dfrac{2-x_2}{2}\right)}{3\left(\dfrac{x_2-3}{3}\right)}\\ \dfrac{dx_1}{dx_2}=\dfrac{2(2-x_2)}{x_2-3}\\ x_1=\int\dfrac{2(2-x_2)}{x_2-3}dx_2 \end{eqnarray}

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    $\begingroup$ Have you tried dividing the two equations to eliminate $dt$? Find $x_1$ in terms of $x_2$ first (or the other way round) and then plug them in their original equations. $\endgroup$ – Shubham Johri Feb 16 at 6:27
  • $\begingroup$ please see my edited question, I have trouble $\endgroup$ – Ongky Denny Wijaya Feb 16 at 6:35
  • $\begingroup$ You may try to express the coefficient of $dt,x_1(1-x_2)$ as a function of $x_1+x_2$ (as the $LHS$ is $d(x_1+x_2)$), then substitute $x_1+x_2=y$ and try to solve the resulting differential equation in $y,t$ $\endgroup$ – Shubham Johri Feb 16 at 6:37
  • $\begingroup$ Please see my edited question, if I tried to dividing 2 equations above, the result is that. It is correct or incorrect? $\endgroup$ – Ongky Denny Wijaya Feb 16 at 6:53
  • $\begingroup$ It is correct. You missed out a $2$ in the numerator in $RHS$. You may integrate both sides from $t=0$ to $t=t$ $\endgroup$ – Shubham Johri Feb 16 at 6:54

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