Problem 6 - Applied Calculus

Question

Find particular solutions \(\frac{d m}{d t}=0.1 m+200, \quad m(0)=1000\)

Step-by-Step Solution

Answer

The particular solution is \( m(t) = 3000e^{0.1t} - 2000 \).

Step 1: Identify the Type of Differential Equation

The given differential equation \( \frac{d m}{d t} = 0.1m + 200 \) is a first-order linear differential equation. It has the form \( \frac{d m}{d t} = am + b \) where \( a = 0.1 \) and \( b = 200 \).

Step 2: Solve the Homogeneous Equation

The homogeneous part of the equation is \( \frac{d m}{d t} = 0.1m \). Solving it involves separation of variables: \( \frac{d m}{m} = 0.1 dt \). Integrating both sides, we get \( \ln |m| = 0.1t + C \), or \( m = Ce^{0.1t} \).

Step 3: Find Particular Solution of the Non-Homogeneous Equation

We use the method of undetermined coefficients. Assume a particular solution \( m_p(t) = A \), a constant. Substitute into \( \frac{dm}{dt} = 0.1m + 200 \): \( 0 = 0.1A + 200 \). Solving for \( A \), we get \( A = -2000 \). Thus, a particular solution is \( m_p(t) = -2000 \).

Step 4: Write the General Solution

The general solution is the sum of the homogeneous solution and the particular solution: \( m(t) = Ce^{0.1t} - 2000 \).

Step 5: Apply Initial Condition

The initial condition is \( m(0) = 1000 \), so substitute into the general solution: \( 1000 = Ce^0 - 2000 \). This simplifies to \( 1000 = C - 2000 \), giving \( C = 3000 \).

Step 6: Write the Particular Solution

Substitute \( C = 3000 \) back into the general solution to get the particular solution: \( m(t) = 3000e^{0.1t} - 2000 \).

Key Concepts

Particular Solution

A particular solution is a specific solution to a differential equation that satisfies the non-homogeneous part of the equation. In the context of our problem, we want to find a solution to the differential equation \( \frac{dm}{dt} = 0.1m + 200 \). This equation includes a constant term (200), which makes it non-homogeneous.

To find a particular solution, we look for one specific function, \( m_p(t) \), that satisfies the entire differential equation. In this case, we assume a simple form for \( m_p(t) \), such as an arbitrary constant \( A \), since we are dealing with linear and constant coefficients. We substitute \( m_p(t) = A \) into the equation. By doing so, we set the derivative to zero in the equation, which becomes \( 0 = 0.1A + 200 \). Solving for \( A \), we find that \( A = -2000 \).

Thus, the particular solution is \( m_p(t) = -2000 \). It's important to note that this particular solution is valid for this specific form of the equation. It allows us to deal with the non-homogeneous part of the differential equation.

Homogeneous Solution

A homogeneous solution focuses on the part of the differential equation without any constant terms, such as \( b \) in \( am + b \) equations. For our equation \( \frac{dm}{dt} = 0.1m + 200 \), the homogeneous counterpart disregards the constant 200. Thus, we consider \( \frac{dm}{dt} = 0.1m \).

To solve it, we employ the separation of variables. This involves moving all terms involving \( m \) to one side and \( t \) to the other, resulting in \( \frac{dm}{m} = 0.1 \, dt \). By integrating both sides, we obtain \( \ln |m| = 0.1t + C \), where \( C \) is an integration constant. Exponentiating both sides gives us \( m = Ce^{0.1t} \).

This solution represents the homogeneous part because it handles the equation without the influence of the external force or constant, \( 200 \). The value of \( C \) is determined after combining with the particular solution and applying initial conditions.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique to find the particular solution of a non-homogeneous linear differential equation. It's especially useful when the non-homogeneous part of the equation is a simple function, such as a constant, polynomial, exponential, or sinusoidal function.

In our exercise, \( \frac{dm}{dt} = 0.1m + 200 \), the non-homogeneous part is the constant term 200. We hypothesize a straightforward form for the particular solution, like a constant \( A \). This is because constant terms in the equation suggest a simple constant particular solution.

After substituting \( m_p(t) = A \) into the differential equation, we equate \( 0.1A + 200 = 0 \) to solve for \( A \). This results in \( A = -2000 \). The strength of this method lies in its simplicity and efficiency when dealing with certain non-homogeneous terms. It saves on computational effort, especially compared to more general methods, by making educated guesses on the form of the particular solution.

Using undetermined coefficients gives us a quick route to finding \( m_p(t) \), helping complete the full solution, and wrapping it into a neat solution in combination with the homogeneous part.