# Autonomous ODE

by williamrand1
Tags: autonomous
 P: 21 Hi everyone, Im looking for an autonomous first order ode that has the following properties. For dependent variable x: x(t=∞)=0 x(t=-∞)=0 and the function x(t) has one maximum. Any help would be great. Rgds...
 P: 681 This is impossible. Suppose the maximum is at $t = t_0$. Then there exist $t_1 < t_0 < t_2$ such that $x(t_1) = x(t_2)$, but $\dot x(t_1) = -\dot x(t_2)$. There is no way to express that requirement in an autonomous first order ODE. You are going to need a second-order autonomous ODE, as should be obvious from the fact that you want to satisfy two boundary conditions.
 P: 743 What do you think of y' = - y^(3/2) ?
P: 21

## Autonomous ODE

 Quote by pasmith This is impossible. Suppose the maximum is at $t = t_0$. Then there exist $t_1 < t_0 < t_2$ such that $x(t_1) = x(t_2)$, but $\dot x(t_1) = -\dot x(t_2)$. There is no way to express that requirement in an autonomous first order ODE. You are going to need a second-order autonomous ODE, as should be obvious from the fact that you want to satisfy two boundary conditions.
Thanks pasmith

Could you explain why it is not possible?
P: 21
 Quote by JJacquelin What do you think of y' = - y^(3/2) ?
Thanks JJ

Is there an exact solution to this?
 P: 743 dy/dx = -y^(3/2) dx = - dy/y^(3/2) x = (2 / y^(1/2)) +C y^(1/2) = 2/(x-C) y = 4/(x-C)²
HW Helper
P: 1,391
 Quote by JJacquelin dy/dx = -y^(3/2) dx = - dy/y^(3/2) x = (2 / y^(1/2)) +C y^(1/2) = 2/(x-C) y = 4/(x-C)²
That has a divergence, not a maximum, though! I'm not sure that's what williamrand1 is looking for.

williamrand1, what about trying to take a function that you know has the properties you desire, differentiate it, and then see if you can rewrite the derivative in terms of x(t), with no explicit time dependence?
 P: 743 Hi williamrand1 ! Then, what about this one : y' = -2y*sqrt(ln(1/y)) which solution is : y = exp(-(x+c)²)

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