Chapter 2: Problem 6
For each initial value problem, (a) Find the general solution of the differential equation. (b) Impose the initial condition to obtain the solution of the initial value problem. \(y^{\prime}-2 y=e^{3 t}, \quad y(0)=3\)
Short Answer
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Question: Solve the following initial value problem: \(y^{\prime}-2 y = e^{3 t}\) with the initial condition \(y(0) = 3\).
Answer: The specific solution for the given initial value problem is \(y(t) = e^{2t}(e^{t} + 2)\).
Step by step solution
01
Solving the Differential Equation
Given the differential equation \(y^{\prime}-2 y = e^{3 t}\), we can rearrange it as follows:
\(y^{\prime}-2 y - e^{3 t} = 0\)
To obtain the general solution of this first-order linear differential equation, we will employ the integrating factor method. The integrating factor is given by \(I(t) = e^{\int(-2) dt} = e^{-2t}\). Multiplying both sides of the equation by the integrating factor, we get:
\(e^{-2t} (y^{\prime}-2 y) = e^{-2t} e^{3t}\)
Now, the left side of the equation resembles the derivative of the product \(y(t)e^{-2t}\):
\(\frac{d(y(t)e^{-2t})}{dt} = e^{t}\)
Now, we integrate both sides with respect to \(t\):
\(\int\frac{d(y(t)e^{-2t})}{dt} dt = \int e^{t} dt\)
\(y(t)e^{-2t} = e^{t} + C\)
Where \(C\) is the constant of integration. Solving for \(y(t)\), we have:
\(y(t) = e^{2t}(e^{t} + C)\)
02
Applying the Initial Condition
Now, we'll use the initial condition \(y(0) = 3\) to find the specific solution of the initial value problem. Substitute \(t = 0\) and \(y = 3\) into the general solution:
\(3 = e^{2\cdot0}(e^{0} + C)\)
\(3 = 1(1 + C)\)
Subtracting 1 from both sides, we get the constant of integration:
\(C = 2\)
Now that we have found the constant of integration, we can provide the specific solution for the initial value problem:
\(y(t) = e^{2t}(e^{t} + 2)\)
This is the solution for the given initial value problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
General Solution of Differential Equations
In the realm of calculus, differential equations play a pivotal role in modeling various physical systems, such as population growth, chemical reactions, and even finance. The general solution of a differential equation represents a family of functions that include every possible particular solution to the differential equation. To find the general solution, one often looks for a function without any fixed values assigned to constants. These constants are determined later by initial or boundary conditions.
When working with first-order linear differential equations, the general solution typically includes an arbitrary constant, denoted by 'C'. This reflects that there's not just one, but an infinite number of functions that can satisfy the equation, each corresponding to a different value of 'C'. The process involves finding an expression for the dependent variable (usually 'y' or 'u') involving one or more arbitrary constants.
When working with first-order linear differential equations, the general solution typically includes an arbitrary constant, denoted by 'C'. This reflects that there's not just one, but an infinite number of functions that can satisfy the equation, each corresponding to a different value of 'C'. The process involves finding an expression for the dependent variable (usually 'y' or 'u') involving one or more arbitrary constants.
Applying Initial Conditions
Once a general solution is obtained, the next step is to convert it into a specific solution that satisfies a given set of initial conditions. These conditions specify the value of the function and possibly its derivatives at a particular point. This step is crucial for tailoring the solution to a particular scenario or problem.
To apply initial conditions, you plug in the given values for the independent variable (usually 't' for time) and the dependent variable into the general solution. This allows you to solve for the arbitrary constants. For example, if an initial condition is given as 'y(0) = 3', you will set the independent variable to 0 and the dependent variable to 3 in the general solution, then solve for 'C'. The resulting equation, now void of any arbitrary constants, represents the specific solution to the differential equation for the initial condition provided.
To apply initial conditions, you plug in the given values for the independent variable (usually 't' for time) and the dependent variable into the general solution. This allows you to solve for the arbitrary constants. For example, if an initial condition is given as 'y(0) = 3', you will set the independent variable to 0 and the dependent variable to 3 in the general solution, then solve for 'C'. The resulting equation, now void of any arbitrary constants, represents the specific solution to the differential equation for the initial condition provided.
Integrating Factor Method
A powerful technique for solving first-order linear differential equations is the integrating factor method. It leverages the concept of an integrating factor, a function that, when multiplied by the original differential equation, allows it to be rewritten in a form that is easily integrable.
The integrating factor is typically denoted as 'I(t)' and is calculated using the integral of the coefficient of 'y' from the standard form of the equation. Once determined, both sides of the differential equation are multiplied by this factor. The left side typically becomes the derivative of the product of the integrating factor and 'y', which simplifies the process as you can then integrate both sides with respect to 't'.
The magic of the integrating factor lies in its ability to transform a seemingly complicated differential equation into a simpler form where standard integration techniques apply, thus paving the way for finding the general solution.
The integrating factor is typically denoted as 'I(t)' and is calculated using the integral of the coefficient of 'y' from the standard form of the equation. Once determined, both sides of the differential equation are multiplied by this factor. The left side typically becomes the derivative of the product of the integrating factor and 'y', which simplifies the process as you can then integrate both sides with respect to 't'.
The magic of the integrating factor lies in its ability to transform a seemingly complicated differential equation into a simpler form where standard integration techniques apply, thus paving the way for finding the general solution.
First-Order Linear Differential Equations
Frequently encountered in various scientific and engineering fields, first-order linear differential equations form a subclass of differential equations that can be expressed as 'y' plus some function of the independent variable times 'y' equals another function of the independent variable. In mathematical form, they're often noted as:\[\begin{equation}\frac{dy}{dt} + p(t) y = g(t)\end{equation}\]where 'p(t)' and 'g(t)' are functions of 't' alone. The term 'first-order' refers to the highest derivative of 'y' present being the first derivative, while 'linear' denotes that 'y' and its derivative appear to no higher than the first power.
The methods for solving these equations, such as the integrating factor technique, rely on their unique structure. This structure is designed to facilitate the search for a solution even when the equation may seem daunting at first glance. By harnessing these systematic approaches, solving for 'y' becomes a much more manageable task.
The methods for solving these equations, such as the integrating factor technique, rely on their unique structure. This structure is designed to facilitate the search for a solution even when the equation may seem daunting at first glance. By harnessing these systematic approaches, solving for 'y' becomes a much more manageable task.