Solving a Second-Order Differential Equation in Physics: A Comprehensive Guide
In physics, differential equations are often used to describe the behavior of systems over time or space. One common example is the second-order differential equation:
(frac{d^2x}{dt^2} -frac{k}{x^2})
This equation describes the motion of a particle under a force proportional to the inverse square of its distance from the origin, often seen in scenarios like gravitational or electrostatic forces.
Step-by-Step Solution
To solve this differential equation, we start by introducing a substitution to simplify the expression. Let's denote v (frac{dx}{dt}). Then, we have:
[frac{d^2x}{dt^2} frac{dv}{dt} frac{dv}{dx} frac{dx}{dt} v frac{dv}{dx}]Substituting this into the original equation gives:
[v frac{dv}{dx} -frac{k}{x^2}]We can now separate the variables:
[v dv -frac{k}{x^2} dx]Integrating both sides, we get:
[int v dv int -frac{k}{x^2} dx]This leads to:
[frac{v^2}{2} -frac{k}{x} C]where C is the constant of integration. Rearranging this equation, we obtain:
[v^2 2left(-frac{k}{x} Cright)]Since (v frac{dx}{dt}), we can write:
[left(frac{dx}{dt}right)^2 2left(-frac{k}{x} Cright)]Taking the square root, we get:
[frac{dx}{dt} pm sqrt{2left(-frac{k}{x} Cright)}]Now we can separate variables again:
[frac{dx}{sqrt{2left(-frac{k}{x} Cright)}} dt]Integrating both sides can be complex and may require substitution techniques or numerical methods for specific values. However, this integral can generally be solved for given initial conditions and values of k and C.
Alternative Approach
Alternatively, let (v) be the velocity of the moving body. Then, from the definitions of velocity and acceleration, we have:
v(frac{dx}{dt}) x(frac{dv}{dt})Applying the chain rule for derivatives, we get:
[x frac{dv}{dt} frac{dv}{dx} cdot frac{dx}{dt} v cdot frac{dv}{dx}]Thus, the original equation can be reformulated as:
[v frac{dv}{dx} -frac{k}{x^2}]Hence:
[v dv -frac{k}{x^2} dx]Integrating both sides of the equation, we obtain:
[frac{v^2}{2} -frac{k}{x} C]Assuming the velocity is zero at a very long distance (x rightarrow infty), then from the equation, C must be zero. Leading to:
[frac{v^2}{2} -frac{k}{x}]thus:
[v^2 -frac{2k}{x}]Considering the positive direction of the vector velocity is the same as the direction of the acceleration, we take the negative value:
[frac{dx}{dt} -sqrt{-frac{2k}{x}}]In order to simplify the problem, we have (x_0), so:
[frac{dx}{dt} -sqrt{frac{2k}{x}}]
This can be further integrated as:
[sqrt{x} dx -sqrt{2k} dt]
[frac{2}{3}x^{frac{3}{2}} -sqrt{2k} t K]
where K is a constant. Using the initial condition x_{t0} d, we get:
[0 -frac{2}{3}d^{frac{3}{2}} K Rightarrow K frac{2}{3}d^{frac{3}{2}}]
Therefore:
[frac{2}{3}x^{frac{3}{2}} -sqrt{2k} t frac{2}{3}d^{frac{3}{2}}]
Finally:
[x left( -frac{3}{2}sqrt{2k} t frac{2}{3}d^{frac{3}{2}} right)^{frac{2}{3}}]
The time for the body to come over distance d can be computed by setting x 0 and solving for t:
[0 left(-frac{3}{2}sqrt{2k} t frac{2}{3}d^{frac{3}{2}}right)^{frac{2}{3}}]
[Rightarrow t frac{2}{3}frac{d^{frac{3}{2}}}{sqrt{2k}}]
Conclusion
The solution to the differential equation (frac{d^2x}{dt^2} -frac{k}{x^2}) involves integrating an expression. Depending on the initial conditions and the values of k and C, the specific form of the solution will vary. Methods like variable separation, initial condition application, and numerical solutions can be used to find the exact form of the solution.