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Find the general solution. \(y^{\prime}+2 y=e^{-2 t}\)

Short Answer

Expert verified
Answer: The general solution of the given differential equation is \(y(t) = Ce^{-2t} + te^{-2t}\).

Step by step solution

01

Solve the homogeneous equation

To find the complementary function, we need to solve the homogeneous equation: \(y' + 2y = 0\). This is a first-order linear homogeneous differential equation, so we can solve it using an integrating factor, which is given by \(e^{\int 2\, dt} = e^{2t}\). Multiply both sides of the equation by the integrating factor: \(e^{2t}y' + 2e^{2t}y = 0\). Now, the left side of this equation is the derivative of the product of \(e^{2t}\) and \(y\), i.e., \((e^{2t}y)'\). Thus, we can write the equation as: \((e^{2t}y)' = 0\). Next, integrate both sides with respect to \(t\): \(\int (e^{2t}y)' dt = \int 0 \, dt\). This gives us: \(e^{2t}y = C\), where \(C\) is the constant of integration. To get the complementary function, we need to express \(y\) in terms of \(t\). So, \(y(t) = Ce^{-2t}\).
02

Find the particular solution

Now, we will find the particular solution of the given non-homogeneous equation: \(y' + 2y = e^{-2t}\). We assume the particular solution to be of the form \(y_p(t) = Ate^{-2t}\), where \(A\) is a constant. Now, we differentiate \(y_p(t)\) to get \(y_p'(t)\): \(y_p'(t) = A(-2te^{-2t} + e^{-2t})\). Then, we substitute \(y_p(t)\) and \(y_p'(t)\) into the non-homogeneous equation: \(A(-2te^{-2t} + e^{-2t}) + 2(At)e^{-2t} = e^{-2t}\). Now, we simplify the equation and solve for \(A\): \(A(-2t + 1)e^{-2t} + A(2t)e^{-2t} = e^{-2t}\). \(Ae^{-2t} = e^{-2t} \implies A = 1\). So, the particular solution is \(y_p(t) = te^{-2t}\).
03

Combine the complementary and particular solutions

Finally, we can find the general solution by combining the complementary function and the particular solution: \(y(t) = y_c(t) + y_p(t) = Ce^{-2t} + te^{-2t}\). The general solution of the given differential equation is: \(y(t) = Ce^{-2t} + te^{-2t}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First-order Differential Equation
A first-order differential equation is a type of equation that involves the first derivative of a function. These equations are essential because they describe how a quantity changes with respect to one another. In mathematical terms, it can be expressed as \[ y' + p(t)y = g(t) \] where \(y'\) is the first derivative of \(y\), and \(p(t)\) and \(g(t)\) are functions of \(t\).

Such equations are commonly used in various fields of science and engineering, as they are fundamental in modeling growth rates, cooling laws, and more. To solve these equations, we identify whether they are linear or non-linear, homogeneous or non-homogeneous, and choose appropriate methods accordingly.

Understanding the structure of first-order differential equations helps in applying methods, such as integrating factors, to find solutions efficiently.
Integrating Factor
The integrating factor is a powerful technique used to solve first-order linear differential equations. Its purpose is to make the left-hand side of the equation an exact derivative, simplifying the integration process.

To determine the integrating factor, we use the formula:\[ \mu(t) = e^{\int p(t) \, dt} \]where \(p(t)\) is the coefficient of \(y\) in the differential equation.

Multiplying the entire differential equation by this integrating factor transforms it such that the left-hand side becomes the derivative of a product of two functions.
  • Step 1: Compute \(\mu(t)\) using \(p(t)\).
  • Step 2: Multiply the entire equation by \(\mu(t)\).
  • Step 3: Integrate both sides to find the solution.
This approach is effective because it reduces the problem to finding the derivative of a single term, making it easier to integrate and solve for \(y(t)\).
General Solution
In the context of differential equations, the general solution represents the complete set of solutions that satisfy the equation. It includes both the complementary solution and particular solution, encompassing all possible behaviors of the system.

For a non-homogeneous first-order differential equation like \[ y' + p(t)y = g(t) \], we
  • First find the complementary solution \(y_c(t)\) by solving the associated homogeneous equation \(y' + p(t)y = 0\).
  • Then, determine a particular solution \(y_p(t)\) that solves the entire non-homogeneous equation.
  • Finally, combine them: \(y(t) = y_c(t) + y_p(t)\).
The general solution provides a comprehensive view of the system, incorporating constant terms that can be adjusted according to initial conditions or additional constraints.
Homogeneous Equation
A homogeneous equation is a special type of differential equation where the function and its derivatives add up to zero. This can be simply described as: \[ y' + p(t)y = 0 \].

Because there is no external input or "forcing function" on the right side of the equation, it models natural, undisturbed behavior of the system.

To solve a homogeneous equation, we typically use an integrating factor or separation of variables. Solving it provides the complementary function \(y_c(t)\), which denotes the intrinsic dynamics of the system without any external influence.

Understanding homogeneous equations helps in studying the natural equilibrium and stability of systems and provides the base solution that combines with particular solutions for non-homogeneous problems.

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Most popular questions from this chapter

When modeling the action of drag chutes and parachutes, we have assumed that the chute opens instantaneously. Real devices take a short amount of time to fully open and deploy. In this exercise, we try to assess the importance of this distinction. Consider again the assumptions of Exercise 2 . A 3000 -lb dragster is moving on a straight track at a speed of \(220 \mathrm{mph}\) when, at time \(t=0\), the drag chute is opened. If we assume that the drag force is proportional to velocity and that the chute opens instantaneously, the differential equation to solve is \(m v^{\prime}=-k v\). If we assume a short deployment time to open the chute, a reasonable differential equation might be \(m v^{\prime}=-k(\tanh t) v\). Since \(\tanh (0)=0\) and \(\tanh (1) \approx 0.76\), it will take about \(1 \mathrm{sec}\) for the chute to become \(76 \%\) deployed in this model. Assume \(k=25 \mathrm{lb}-\mathrm{sec} / \mathrm{ft}\). Solve the two differential equations and determine in each case how long it takes the vehicle to slow to \(50 \mathrm{mph}\). Which time do you anticipate will be larger? (Explain.) Is the idealization of instantaneous chute deployment realistic?

A tank originally contains 100 gal of fresh water. At time \(t=0\), a solution containing \(0.2 \mathrm{lb}\) of salt per gallon begins to flow into the tank at a rate of \(3 \mathrm{gal} / \mathrm{min}\) and the well-stirred mixture flows out of the tank at the same rate. (a) How much salt is in the tank after \(10 \mathrm{~min}\) ? (b) Does the amount of salt approach a limiting value as time increases? If so, what is this limiting value and what is the limiting concentration?

Use the ideas of Exercise 32 to solve the given initial value problem. Obtain an explicit solution if possible. $$ y^{\prime}=\frac{y+t}{y+t+1}, \quad y(-1)=0 $$

On August 24,1894 , Pop Shriver of the Chicago White Stockings caught a baseball dropped (by Clark Griffith) from the top of the Washington Monument. The Washington Monument is \(555 \mathrm{ft}\) tall and a baseball weighs \(5 \frac{1}{8} \mathrm{oz}\). (a) If we ignore air resistance and assume the baseball was acted upon only by gravity, how fast would the baseball have been traveling when it was \(7 \mathrm{ft}\) above the ground? (b) Suppose we now include air resistance in our model, assuming that the drag force is proportional to velocity with a drag coefficient \(k=0.0018 \mathrm{lb}-\mathrm{sec} / \mathrm{ft}\). How fast is the baseball traveling in this case when it is \(7 \mathrm{ft}\) above the ground?

In each exercise, the general solution of the differential equation \(y^{\prime}+p(t) y=g(t)\) is given, where \(C\) is an arbitrary constant. Determine the functions \(p(t)\) and \(g(t)\). \(y(t)=C t^{-1}+1, \quad t>0\)

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