Solve ODE in Plane: d2x/dt2=-cosx, d2y/dt2=-cosy

• wofsy
In summary, you can solve the ODE, "d2x/dt2=-cosx d2y/dt2=-cosy", by integrating each equation twice.

wofsy

Can someone tell me how to solve the ODE,

d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?

You can use separation of variables and then integrate twice (with an integration constant!)

thanks

Those two equations are completely independent- so it is not necessary to "separate variables". You are just solving two separate second order differential equations. And, in fact you just need to integrate each twice.

HallsofIvy said:
Those two equations are completely independent- so it is not necessary to "separate variables". You are just solving two separate second order differential equations. And, in fact you just need to integrate each twice.

Hrm? They don't look independent to me. The derivatives are with respect to t, not x or y. Integrating the x equation would give, for instance,

$$\dot{x(t)} - \dot{x(t_0)} = \int_{t_0}^{t}d\tau~\cos y(\tau)$$

which isn't so useful if you can't solve for what y(t) is.

If d2x/dt2= cos(x) is a single equation in the dependent variable x as a function of t. There is absolutely no reason to introduce y. Since the independent variable "t" does not appear in the equation, I would use "quadrature":

Let v= dx/dt so that d2x/dt2= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v2= sin(x)+ C. dx/dt= v= $\sqrt{2(sin(x)+ C)}$ or
$$\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt$$
That left side is an "elliptical integral".

Of course, y will be exactly the same, though possibly with different constants of integration.

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HallsofIvy said:
If d2x/dt2= cos(x) is a single equation in the dependent variable x as a function of t. There is absolutely no reason to introduce y. Since the independent variable "t" does not appear in the equation, I would use "quadrature":

Let v= dx/dt so that d2x/dt2= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v2= sin(x)+ C. dx/dt= v= $\sqrt{2(sin(x)+ C)}$ or
$$\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt$$
That left side is an "elliptical integral".

Of course, y will be exactly the same, though possibly with different constants of integration.

Ah, I see, I didn't parse the problem the way it was intended to be read. I read it as

$$\frac{d^2x}{dt^2} = -\cos x \frac{d^2y}{dt^2} = -\cos y$$

i.e.,

$$\frac{d^2x}{dt^2} = -\cos y$$
and
$$\cos x \frac{d^2y}{dt^2} = \cos y$$

This is why I always use some sort of punctuation in between separate equations written on the same line. =P

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And how do you know that was how it was "intended to be read"?

I was able to get to the elliptic integral. But I have no idea what it looks like. Further, if x and y are both the same elliptic integral then the orbits in the plane should be fairly simple. But what do they look like?

1. What is an ODE in plane?

An ODE (ordinary differential equation) in plane refers to a differential equation that involves two independent variables, usually represented as x and y. The equation can be solved to find the values of x and y that satisfy the equation.

2. What does d2x/dt2=-cosx mean?

This notation means that the second derivative of x with respect to time (t) is equal to the cosine of x. In other words, the rate of change of the rate of change of x is equal to the cosine of x.

3. How do you solve d2x/dt2=-cosx and d2y/dt2=-cosy?

To solve these equations, we need to use techniques from calculus, such as integration and differentiation. We can also use numerical methods or computer software to find approximate solutions.

4. What are the possible solutions for d2x/dt2=-cosx and d2y/dt2=-cosy?

The possible solutions depend on the initial conditions given for x and y. These initial conditions determine the specific values of x and y that satisfy the equations. There can be multiple solutions or no solution at all.

5. Why is solving ODE in plane important in science?

Solving ODEs in plane is important in science because many natural phenomena can be described by these equations. From predicting the motion of planets to modeling chemical reactions, ODEs in plane help us understand and make predictions about the world around us.