Problem 33 - Calculus and its applications

Question

(a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation. \(\frac{d M}{d t}=M,\) where \(M=6\) when \(t=0\)

Step-by-Step Solution

Answer

The particular solution is \( M = 6e^t \) and it satisfies the differential equation.

Step 1: Identify the Differential Equation Type

The differential equation given is \( \frac{dM}{dt} = M \). This is a separable first-order linear differential equation.

Step 2: Separate Variables

To solve the differential equation, separate the variables by dividing both sides by \( M \) and multiplying both sides by \( dt \): \( \frac{1}{M} dM = dt \).

Step 3: Integrate Both Sides

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

Step 4: Solve for M

Exponentiate both sides to solve for \( M \): \( M = e^{t+C} = e^t \cdot e^C \). Let \( e^C = C_1 \), then \( M = C_1 e^t \).

Step 5: Use Initial Condition to Find Particular Solution

The initial condition given is \( M = 6 \) when \( t = 0 \). Substituting these values into the general solution \( M = C_1 e^t \), we have \( 6 = C_1 e^0 \), which simplifies to \( 6 = C_1 \). Therefore, the particular solution is \( M = 6e^t \).

Step 6: Verify the Solution

Substitute \( M = 6e^t \) back into the original differential equation and check: \( \frac{dM}{dt} = \frac{d}{dt}(6e^t) = 6e^t \) and the right-hand side of the original equation is also \( M = 6e^t \). Both sides match, verifying the solution.

Key Concepts

Separable Equations

A separable equation is a type of differential equation that can be written so that all terms involving one variable, like \( M \), are on one side of the equation, while all terms involving the other variable, often \( t \), are on the opposite side. In our example, we start with the differential equation \( \frac{dM}{dt} = M \). To separate variables, we rearrange it as \( \frac{1}{M} dM = dt \). This makes it easy to integrate both sides straightforwardly. Separable equations often involve these essential steps:

• Rearrange the equation to isolate terms of each variable on different sides.
• Integrate both sides to solve for the variable functions.
• Use algebra to solve for the function generally or particularly.

These steps help you identify that each side of the equation corresponds to a specific variable, simplifying the process of integration. By understanding these concepts, you can solve complex problems more efficiently.

First-order Linear Differential Equations

First-order linear differential equations are fundamentally important in both theoretical and applied mathematics. These equations involve derivatives of the first order (hence 'first-order'), and they typically take the form \( \frac{dy}{dx} + P(x)y = Q(x) \). This type of equation is linear because it appears in the form of a linear polynomial. In our example, \( \frac{dM}{dt} = M \) is a simplified version of the general form, with the solution showing exponential behavior.Solving these equations generally involves:

• Recognizing the linear form and using an integrating factor where applicable.
• Multiplying through by an integrating factor to simplify the equation.
• Finding the general solution, often as some function plus a constant.

In some cases where factors can multiply straight through like our example, we find direct solutions quite effectively. Understanding the basic theory behind first-order linear differential equations enhances insight into more complicated topics and solutions.

Initial Value Problems

An initial value problem includes not only the differential equation itself but also a specific condition that the solution must satisfy. This condition is given at a particular 'initial' point, hence the name. In our example, we are given the initial condition \( M = 6 \) when \( t = 0 \). Initial value problems proceed by:

• Solving the differential equation to obtain a general solution.


• Substituting the initial value into the general solution to find specific constants.


• Deriving the particular solution that fits the initial condition.

These problems are especially crucial when solutions model real-world phenomena, providing necessary starting points for prediction and analysis. By confirming the solution with initial conditions, you ensure it properly models the scenario at hand.