Chapter 5: Problem 33
(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 \(t=0\)
Short Answer
Expert verified
The particular solution is \( M = 6e^t \) and it satisfies the differential equation.
Step by step solution
01
Identify the Differential Equation Type
The differential equation given is \( \frac{dM}{dt} = M \). This is a separable first-order linear differential equation.
02
Separate Variables
To solve the differential equation, separate the variables by dividing both sides by \( M \) and multiplying both sides by \( dt \): \( \frac{1}{M} dM = dt \).
03
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: \( \ln |M| = t + C \).
04
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 \( M = C_1 e^t \).
05
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 \( M = 6e^t \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.