Problem 35 - Calculus

Question

Solve the differential equations in Problems \(34-43 .\) Assume \(a, b,\) and \(k\) are nonzero constants. $$\frac{d Q}{d t}-\frac{Q}{k}=0$$

Step-by-Step Solution

Answer

The solution is \( Q(t) = C' e^{\frac{t}{k}} \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d Q}{d t} - \frac{Q}{k} = 0 \). This is a first-order linear homogeneous ordinary differential equation (ODE).

Step 2: Rearrange the Equation

Rearrange the equation to make it easier to solve: \( \frac{d Q}{d t} = \frac{Q}{k} \). This format highlights the separation of variables needed for the next step.

Step 3: Separate the Variables

Separate the variables by dividing both sides by \( Q \) and multiplying by \( dt \): \( \frac{dQ}{Q} = \frac{1}{k} \, dt \).

Step 4: Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{dQ}{Q} = \int \frac{1}{k} \, dt \). The left side integrates to \( \ln |Q| \) and the right side integrates to \( \frac{t}{k} + C \), where \( C \) is the constant of integration.

Step 5: Solve for Q

Solve for \( Q \) by exponentiating both sides to eliminate the natural logarithm: \( Q = e^{\frac{t}{k} + C} = e^{\frac{t}{k}} \cdot e^C \). Define \( e^C \) as a new constant, \( C' \), so \( Q = C' e^{\frac{t}{k}} \).

Step 6: Finalize the Solution

The solution of the differential equation is \( Q(t) = C' e^{\frac{t}{k}} \), where \( C' \) is an arbitrary constant that can be determined from initial conditions.

Key Concepts

First-order Differential Equations

A first-order differential equation is an equation involving a function and its first derivative. It has the general form: \[ \frac{dy}{dx} = f(x, y) \]
Here, the equation relates the rate of change of a quantity (the derivative \( \frac{dy}{dx} \)) to the quantity itself and possibly the independent variable \( x \). It's called "first-order" because it involves the first derivative.
These equations are fundamental in modeling phenomena where the change of a system is proportional to the state of the system. Some real-life examples include population dynamics, and temperature change in a rod. Understanding these models can help predict future behavior based on initial conditions.

Separation of Variables

Separation of variables is a systematic method for solving first-order differential equations. It's especially useful when both sides of the equation can be written as a product of a function of \( x \) and a function of \( y \). This allows for the equation to be reorganized so all \( x \)-terms are on one side and all \( y \)-terms are on the other.
The main goal of this technique is to make the equation ready for integration. Let's consider the differential equation \( \frac{dy}{dx} = g(x)h(y) \). To separate variables, it can be rewritten as: \[ \frac{1}{h(y)} \, dy = g(x) \, dx \]
This step simplifies solving the equation because now each side can be integrated individually, leading to the solution of the differential equation.

Integration

Integration is a process of finding the original function given its derivative. It is the inverse operation to differentiation. For differential equations, integration is the step where the separated variables are solved. In the context of separation of variables, after rearranging the equation, each side is integrated:

• For \( \int \frac{dQ}{Q} \), the integral evaluates to \( \ln|Q| \).
• For \( \int \frac{1}{k}dt \), the result is \( \frac{t}{k} + C \).

In integration, constants often appear, denoted simply as \( C \), representing the range of possible solutions. These constants are determined based on given initial conditions. Integration transforms the problem into one that is solvable, providing the exact function or relationship we are trying to find.

Homogeneous Equations

A homogeneous differential equation is one where every term is a multiple of the unknown function or its derivatives. In other words, if the equation can be expressed such that every summand involves the dependent variable (or its derivatives), it is homogeneous.
The simplest example is the linear homogeneous equation \( f'(x) = k \cdot f(x) \). Solving these equations results in solutions where the function and its derivatives can be evenly scaled.
Such functions often describe systems in equilibrium or systems that return to equilibrium over time. In the given exercise, the equation \( \frac{dQ}{dt} - \frac{Q}{k} = 0 \) is homogeneous because simplifying the equation doesn't involve any constant free-term on its own. It characterizes scenarios with proportional growth or decay.