Calculating Equilibrium Points for a Second Order Equation

[franky2727]

ive been doing Xdot and Ydot questions using the formula lamda²-(a+d)lamda +ad-bc=0 where Xdot=ax+bc and ydot=cx+dy no problems so far until this question where i have the second order equation xdot dot+x-1/4x³=0

so i set Xdot dot =Ydot and therefore Xdot =y as usual but this gives me Ydot=1/4x³ -x which obviously doesn't fit the formula above, how do i calculate the equlibrium points in this case

 

[Defennder]

This is a special type of 2nd order ODE. If you have x'' = f(x), you can use the chain rule to express it in the form:

\frac{1}{2} \frac{d}{dx} \left( \frac{dx}{dt} \right)^2 = f(x) which is separable.

 

[Dick]

ive been doing Xdot and Ydot questions using the formula lamda²-(a+d)lamda +ad-bc=0 where Xdot=ax+bc and ydot=cx+dy no problems so far until this question where i have the second order equation xdot dot+x-1/4x³=0

so i set Xdot dot =Ydot and therefore Xdot =y as usual but this gives me Ydot=1/4x³ -x which obviously doesn't fit the formula above, how do i calculate the equlibrium points in this case

The first type of problems are second order linear ODE's with constant coefficients in two variables. The other one is not that type. To find equilibrium points just set x''(t)=x'(t)=0 and solve for x(t).

 

[franky2727]

what is x''t and x't in terms of Xdot?

 

[Dick]

what is x''t and x't in terms of Xdot?

x'(t)=Xdot=dx/dt and x''(t)=Xdot dot=d^2x/dt^2.

 

[franky2727]

giving me 4x-xcubed=0 giving me x=0 2 or -2 ?

 

[Dick]

giving me 4x-xcubed=0 giving me x=0 2 or -2 ?

Sure.

 

[franky2727]

cool chears