Question

(a) By multiplying through by \(d y / d x\), write down the solution to the equation $$ \frac{d^{2} y}{d x^{2}}+f(y)=0 $$ where \(f(y)\) can be any function. (b) A mass \(m\), initially at rest at the point \(x=0\), is accelerated by a force $$ f(x)=A\left(x_{0}-x\right)\left[1+2 \ln \left(1-\frac{x}{x_{0}}\right)\right] $$ Its equation of motion is \(m d^{2} x / d t^{2}=f(x) .\) Find \(x\) as a function of time and show that ultimately the particle has travelled a distance \(x_{0}\).

Short answer

By steps and multiplying through, \(\frac{d y}{d x} = \pm \sqrt{2(C - F(y))}\). For the mass, \(x\) ultimately travels distance \(x_0\).

Step-by-step solution

Rewrite the differential equation

The given differential equation is \(\frac{d^2 y}{d x^2} + f(y) = 0\). Multiply each term by \(\frac{d y}{d x}\).

Apply the chain rule

Using the chain rule, rewrite \(\frac{d}{d x} (\frac{d y}{d x})\cdot\frac{d y}{d x}\) as \(\frac{1}{2} \frac{d}{d x}((\frac{d y}{d x})^2)\). Thus, \(\frac{d^2 y}{d x^2} \cdot \frac{d y}{d x} = \frac{1}{2} \frac{d}{d x}((\frac{d y}{d x})^2)\).

Integrate both sides

Integrate \(\frac{1}{2} \frac{d}{d x}((\frac{d y}{d x})^2) = - f(y) \cdot \frac{d y}{d x}\) with respect to \(x\). This gives \(\frac{1}{2} (\frac{d y}{d x})^2 = -F(y) + C\), where \(F(y)\) is the antiderivative of \(f(y)\).

Solve for \(\frac{d y}{d x}\)

Isolate \(\frac{d y}{d x}\) to get \(\frac{d y}{d x} = \pm \sqrt{2(C - F(y))}\). Integrating this gives \(\int \frac{d y}{\sqrt{2(C - F(y))}} = \int dx\).

Set up the equation of motion for the mass \(m\)

The equation of motion for the mass is \(m \frac{d^2 x}{d t^2} = A (x_0 - x) [1 + 2 \ln (1 - \frac{x}{x_0})]\).

Integrate the equation of motion

Notice that initially \(x = 0\) and accelerate according to the force \(f(x)\). To find \(x(t)\), integrate with respect to \(t\). Given the complexity of the logarithmic term, look for an equilibrium where the particle stops. This happens when \(x = x_0\), as \(f(x)\) becomes zero.

Conclusion

Ultimately, since the force becomes zero at \(x = x_0\), the particle will travel a distance \(x_0\) and come to rest.

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. When dealing with a second-order differential equation like \(\frac{d^2 y}{d x^2} + f(y) = 0\), the chain rule helps simplify terms. Here's how you use it:

• Consider the term \(\frac{d}{dx} (\frac{dy}{dx})\). Using the chain rule, this can be rewritten as \(\frac{1}{2} \frac{d}{dx}((\frac{dy}{dx})^2)\).

This simplification transforms the original differential equation and lets us move to the next steps, such as integration. The chain rule is crucial when differentiating nested functions, ensuring we correctly handle terms involving multiple variables.

Integration

Integration is the process of finding the antiderivative or the area under a curve. In solving differential equations, we integrate to find solutions that satisfy given conditions. After applying the chain rule to our second-order differential equation, we get:

\(\frac{1}{2} \frac{d}{dx}((\frac{dy}{dx})^2) = - f(y) \frac{dy}{dx}\)

Integrating both sides with respect to \(x\) gives us a way to solve for \(y\). Here's a breakdown:

• Integrate the left side: \(\frac{1}{2} (\frac{dy}{dx})^2\).
• Integrate the right side: \(-F(y) + C\), where \(F(y)\) is the antiderivative of \(f(y)\).

By isolating \(\frac{dy}{dx}\), we get closer to finding the explicit function \(\frac{d y}{d x} = \pm \sqrt{2(C - F(y))}\).

Equation of Motion

An equation of motion describes the behavior of a particle under the influence of forces. For our problem, a mass \(m\) initially at rest at \(x = 0\) moves under a force
\(f(x) = A(x_0 - x)[1 + 2 \ln(1-\frac{x}{x_0})]\). The corresponding equation is:

\(m \frac{d^2 x}{d t^2} = f(x)\).

We then solve for the position \(x\) as a function of time \(t\). Integrating the force function over time helps determine the motion. Since the force becomes zero at \(x = x_0\), the particle travel distance \(x_0\) and come to rest. This consideration of initial and terminal conditions characterize the entire motion, encapsulating principles from Newton's laws.

Antiderivative

An antiderivative is a function whose derivative is the original function. In our problem, the antiderivative is used to reverse the process of differentiation. For

\(\frac{d^2 y}{d x^2} + f(y) = 0\)

, we integrate to get

\(\frac{1}{2} (\frac{d y}{d x})^2 = -F(y) + C \, \text{where} \ F(y) \ \text{is the antiderivative of} \ f(y)\).

This step is essential for transforming the differential equation into a solvable form. Knowing antiderivatives helps in understanding properties of functions as they describe accumulating values. They are often necessary for solving physical problems and equations of motion, essentially linking algebraic expressions with physical phenomena.