Solving 2nd Order ODE: du/ds & k^2

In summary, the conversation discusses the equation d^{2}u/ds^{2}= cosu[(du/ds)^{2} - k^{2}] and how it can be solved using the technique of "quadrature". It also mentions using the variable v= du/ds to simplify the equation and solve it as a separable first order equation. Once v is solved, it can be integrated to find u.
  • #1
lavinia
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d[itex]^{2}[/itex]u/ds[itex]^{2}[/itex]= cosu[(du/ds)[itex]^{2}[/itex] - k[itex]^{2}[/itex]]
 
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  • #2
"cosu" is cos(u)? Since the independent variable, s, does not appear explicitely in that equation you can use a technique called "quadrature".

Let v= du/ds so that [itex]d^2u/ds^2=dv/ds[/itex] but, by the chain rule, dv/ds= (dv/du)(du/ds)= v(dv/du) so your equation becomes [itex]v dv/du= cos(u)(v^2+ k^2)[/itex]

That's a separable first order equation. Once you have solved it for v, integrate to find u.
 
  • #3
lavinia said:
d[itex]^{2}[/itex]u/ds[itex]^{2}[/itex]= cosu[(du/ds)[itex]^{2}[/itex] - k[itex]^{2}[/itex]]

That lends itself to interpretation:

[tex]\frac{d^2 u}{ds^2}=\cos(u(v))\biggr|_{v=(u')^2-k^2}[/tex]

that's doable right?
 
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  • #4
HallsofIvy said:
"cosu" is cos(u)? Since the independent variable, s, does not appear explicitely in that equation you can use a technique called "quadrature".

Let v= du/ds so that [itex]d^2u/ds^2=dv/ds[/itex] but, by the chain rule, dv/ds= (dv/du)(du/ds)= v(dv/du) so your equation becomes [itex]v dv/du= cos(u)(v^2+ k^2)[/itex]

That's a separable first order equation. Once you have solved it for v, integrate to find u.

thanks - pretty cool
 
  • #5


I would say that solving second order ordinary differential equations (ODEs) is an important task in many fields of science, including physics, engineering, and mathematics. In this particular case, we are dealing with an ODE that involves the first and second derivatives of a function u with respect to a variable s, as well as a constant k.

The equation can be rewritten in a more standard form as d^2u/ds^2 + k^2 = cos(u)(du/ds)^2. This type of ODE is known as a second order nonlinear ODE, as it involves both the function u and its derivatives. In general, solving nonlinear ODEs is more challenging than solving linear ODEs, but there are various techniques and methods that can be used to find solutions.

One approach to solving this ODE could be to use a numerical method, such as the Runge-Kutta method, to approximate the solution. This method involves breaking the problem into smaller steps and using iterative calculations to find an approximate solution. Another approach could be to use an analytical method, such as separation of variables or the method of undetermined coefficients, to find a closed-form solution. However, these methods may not always be possible or practical, depending on the complexity of the equation.

In conclusion, solving this second order nonlinear ODE requires careful analysis and the use of appropriate techniques. The solution will depend on the specific values of k and the initial conditions of the problem. As a scientist, it is important to carefully consider the assumptions and limitations of any solution method used and to verify the results through experimentation or comparison with other known solutions.
 

FAQ: Solving 2nd Order ODE: du/ds & k^2

What is a 2nd order ODE?

A 2nd order ODE (Ordinary Differential Equation) is a mathematical equation that involves a dependent variable, its derivatives, and an independent variable. It represents the relationship between the variable and its rate of change.

What is "du/ds" in the context of a 2nd order ODE?

"du/ds" refers to the derivative of the dependent variable with respect to the independent variable in a 2nd order ODE. It represents the rate of change of the dependent variable over a small increment of the independent variable.

What is the significance of "k^2" in a 2nd order ODE?

"k^2" represents a constant in the differential equation. It can affect the behavior of the solution and determine whether it is oscillatory, exponential, or a combination of both.

How do you solve a 2nd order ODE with "du/ds" and "k^2"?

To solve a 2nd order ODE with "du/ds" and "k^2", you can use various methods such as separation of variables, substitution, or integrating factors. These methods involve manipulating the equation to isolate the dependent variable and its derivatives, and then integrating to find the solution.

What are some real-life applications of solving 2nd order ODEs with "du/ds" and "k^2"?

2nd order ODEs with "du/ds" and "k^2" have many real-life applications, such as in physics, engineering, and economics. For example, they can be used to model the motion of a pendulum, the vibrations of a guitar string, or the growth of a population. They can also be used to analyze electrical circuits, chemical reactions, and financial markets.

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