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d[itex]^{2}[/itex]u/ds[itex]^{2}[/itex]= cosu[(du/ds)[itex]^{2}[/itex] - k[itex]^{2}[/itex]]
lavinia said:d[itex]^{2}[/itex]u/ds[itex]^{2}[/itex]= cosu[(du/ds)[itex]^{2}[/itex] - k[itex]^{2}[/itex]]
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.
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.
"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.
"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.
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.
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.