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Can someone tell me how to solve the ODE,

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

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

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

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Can someone tell me how to solve the ODE,

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

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

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

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thanks

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HallsofIvy said:

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,

[tex]\dot{x(t)} - \dot{x(t_0)} = \int_{t_0}^{t}d\tau~\cos y(\tau)[/tex]

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

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If d^{2}x/dt^{2}= 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 d^{2}x/dt^{2}= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v^{2}= sin(x)+ C. dx/dt= v= [itex]\sqrt{2(sin(x)+ C)}[/itex] or

[tex]\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt[/tex]

That left side is an "elliptical integral".

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

Let v= dx/dt so that d

[tex]\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt[/tex]

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:^{2}x/dt^{2}= 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 d^{2}x/dt^{2}= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v^{2}= sin(x)+ C. dx/dt= v= [itex]\sqrt{2(sin(x)+ C)}[/itex] or

[tex]\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt[/tex]

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

[tex]\frac{d^2x}{dt^2} = -\cos x \frac{d^2y}{dt^2} = -\cos y [/tex]

i.e.,

[tex]\frac{d^2x}{dt^2} = -\cos y[/tex]

and

[tex]\cos x \frac{d^2y}{dt^2} = \cos y [/tex]

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"?

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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.

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.

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.

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.

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.

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