Problem 3 - Applied Calculus

Question

Find particular solutions. $$\frac{d H}{d t}=3(H-75), \quad H=0\( when \)t=0$$

Step-by-Step Solution

Answer

The particular solution is \( H = 75 - 75e^{3t} \).

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{dH}{dt} = 3(H - 75) \). This is a first-order linear differential equation that describes a change in \( H \) with respect to time \( t \).

Step 2: Separate Variables

Re-write the equation in a form that allows you to separate variables. We start by dividing both sides by \( H - 75 \), to get \( \frac{dH}{H - 75} = 3 \, dt \).

Step 3: Integrate Both Sides

Integrate both sides of the equation with respect to their variables. This gives us \( \int \frac{dH}{H - 75} = \int 3 \, dt \). The integrals are \( \ln|H - 75| = 3t + C \), where \( C \) is the constant of integration.

Step 4: Solve for H

Exponentiate both sides to remove the natural log: \( |H - 75| = e^{3t + C} \). Let \( e^C = C_1 \), so \( H - 75 = C_1 e^{3t} \) or \( H - 75 = -C_1 e^{3t} \). Choose \( C_1 \) such that it fits any initial conditions.

Step 5: Apply Initial Condition

We are given that \( H = 0 \) when \( t = 0 \). Substitute these into the equation to find \( C_1 \): \( 0 - 75 = C_1 e^{3 \times 0} \). This simplifies to \( -75 = C_1 \), so our particular solution becomes \( H = 75 - 75e^{3t} \).

Step 6: Confirm the Particular Solution

The particular solution that meets the initial condition is \( H = 75 - 75e^{3t} \). You can verify this by substitution back into the original differential equation.

Key Concepts

First-order linear differential equations

First-order linear differential equations are a type of differential equation that involve the first derivative of a function. They are called 'linear' because the function itself and its derivatives appear to the first power, and not higher degrees like squares or cubes. In the given exercise, the equation \(\frac{dH}{dt} = 3(H - 75)\) is in this category. Such equations are useful in modeling systems where a quantity depends on its current value and another variable, like time.

To identify a first-order linear differential equation, look for these characteristics:

• The highest order derivative present is the first derivative \(\frac{dH}{dt}\).
• The functions or their derivatives are not raised to any power other than one.
• Terms can be rearranged to fit the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\), although sometimes rearrangement is not necessary.

Recognizing these equations allows us to apply specific solution techniques, such as separation of variables, making the processes predictable.

Initial value problems

Initial value problems involve solving differential equations with an additional condition called the initial condition. This condition specifies the value of the function at a particular point, allowing for the computation of the constant from the solutions of the equation.

In the exercise, the initial condition is provided by stating that \( H = 0 \) when \( t = 0 \). This initial condition is essential as it helps determine the particular solution from a family of solutions given by indefinite integration. Here's why initial conditions matter:

• They convert a general solution, which includes arbitrary constants, into a specific solution fitting the problem's requirements.
• It matches the solution with real-world scenarios, ensuring that the mathematics reflects actual conditions.

Applying initial conditions is straightforward but crucial, as it ensures that the solution accurately mirrors the situation being modeled.

Separation of variables

Separation of variables is a powerful method for solving first-order differential equations by separating the variables involved. The idea is to isolate all terms containing one variable on one side of the equation and all terms containing the other variable on the opposite side. This method is particularly useful for equations that can be rearranged into the separable form.

In the exercise, the equation \(\frac{dH}{dt} = 3(H - 75)\) was separated into \(\frac{dH}{H - 75} = 3 \, dt\) by dividing both sides by \(H - 75\).

This approach works well when:

• Both sides of the equation can be expressed as functions of their respective variables only.
• The integral of each side can be computed independently.

Through integration and application of algebraic techniques, we solve for the function explicitly. Following separation of variables, the solution progresses through integration and substitution of initial conditions to find a particular solution that fits the equation's constraints.