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Solve differential equation $\frac{dx}{dt}$ = $x (1 − x) \cos (x)$ with $x(t)$ being the solution of the ODE and $x(0) = \frac{1}{2}$ and find limit $t\rightarrow\infty$ of $x(t)$.

I'm having trouble integrating the equation.

I did rewrite the equation to

$$\frac{dt}{dx} = \frac{1}{x(1-x)\cos(x)} = \frac{\sec(x)}{x(1-x)}$$

How do I integrate this to solve the ODE?

MonkeyKing
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Lello
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    Hint: Don't try to solve it. Instead you should observe it's an autonomous equation which mean sketch phase line diagram. – Jacky Chong Sep 28 '16 at 04:58

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I think that you have not to find an explicit solution here. What do you know about existence and uniqueness for a Cauchy problem?

A sketch of the proof.

A unique solution exists for $t>0$. By uniqueness, since $x(0)=1/2\in(0,1)$ and $x=0$ and $x=1$ are stationary solutions, it follows that $x(t)\in (0,1)$ for $t>0$. Moreover $x(t)\in (0,1)$ implies that $x'(t)>0$ which means that the solution is strictly increasing. Therefore it has a limit $L\in (1/2,1]$ at infinity. Try to prove that the limit is $1$ by using Limit of the derivative of a function as x goes to infinity

Community
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Robert Z
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  • Now that you have mention it I read the theorems and sample problems. Is it correct that because x'(t) >0 this means it is increasing function and the limit of t going to infinity is 1/2? Do you mind explaining this step by step because I don't quite grasp the concept yet. Like how can does the equilibrium solution connect to solving this using the theorems? – Lello Sep 28 '16 at 05:34
  • @Lello By uniqueness, since $x(0)\in(0,1)$ then $x(t)\in (0,1)$ for $t>0$. There it is strictly increasing, therefore the limit at infinity exists and the limit is $(1/2,1]$. Try to prove that the limit is $1$. – Robert Z Sep 28 '16 at 06:11