# Second order equation problem

1. Aug 18, 2008

### 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 doesnt fit the formula above, how do i calculate the equlibrium points in this case

2. Aug 18, 2008

### 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.

3. Aug 18, 2008

### Dick

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).

4. Aug 19, 2008

### franky2727

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

5. Aug 19, 2008

### Dick

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

6. Aug 19, 2008

### franky2727

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

7. Aug 19, 2008

### Dick

Sure.

8. Aug 19, 2008

cool chears