/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 (a) find the particular solution... [FREE SOLUTION] | 91Ó°ÊÓ

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(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.
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.

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