The differential equation \(\ddot{x} + k \cos{x} = 0\) can be solved using the method of quadrature. By substituting \(y = \frac{dx}{dt}\), the equation transforms into a first-order equation \(y \frac{dy}{dx} = -\cos(x)\). Integrating this yields \(\frac{1}{2}y^2 = \sin(x) + C\), leading to the expression \(y = \sqrt{2C - 2\sin(x)}\). This approach effectively simplifies the problem, although further integration may be complex.
PREREQUISITESStudents and educators in mathematics, particularly those focusing on differential equations, as well as engineers and physicists dealing with oscillatory systems.
Since the independent variable, which I will call "t", does not appear explicitely in the equation, this is a candidate for "quadrature". Let y= dx/dt. Then d^2x/dt^2= (dy/dx)(dx/dt) by the chain rule. But since y= dx/dt, that is d^2x/dt^2= y dy/dx and the equation converts to the first order equation, for y as a function of x, y dy/dx= -cos(x) which is y dy= -cos(x)dx. Integrating, (1/2)y^2= sin(x)+ C. Now we have y^2= -2 sin(x)+ 2C or y= dx/dt= \sqrt{2C- 2sin(x)} so dx= \sqrt{2C- 2sin(x)}dx.Reid said:Homework Statement
I can not determine the solution to the diff. eq.
Homework Equations
\ddot{x}+k \cos{x}=0.
The constant k is positive.
The Attempt at a Solution
I tried solving it with the methods I know and it all ended up in a big mess. I am just not used to the second term, \cos. Doe's anyone know what to do?