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Chapter 2: Problem 19

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

Assume Newton's law of cooling applies. The temperature of an object is raised from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(10 \mathrm{~min}\) when placed within a \(300^{\circ} \mathrm{F}\) oven. What oven temperature will raise the object's temperature from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(5 \mathrm{~min}\) ?

Suppose that at some initial time the pendulum is located at angle \(\theta_{0}\) with an angular velocity \(d \theta / d t=\omega_{0}\) radians/sec. (a) Equation (15) is a second order differential equation. Rewrite it as a first order separable equation by adopting angle \(\theta\) as the independent variable, using the fact that $$ \theta^{\prime \prime}=\frac{d}{d t}\left(\frac{d \theta}{d t}\right)=\frac{d \omega}{d t}=\frac{d \omega}{d \theta} \frac{d \theta}{d t}=\omega \frac{d \omega}{d \theta} . $$ Complete the specification of the initial value problem by specifying an appropriate initial condition. (b) Obtain the implicit solution $$ m l^{2} \frac{\omega^{2}}{2}-m g l \cos \theta=m l^{2} \frac{\omega_{0}^{2}}{2}-m g l \cos \theta_{0^{-}} $$ The pendulum is a conservative system; that is, energy is neither created nor destroyed. Equation (16) is a statement of conservation of energy. At a position defined by the angle \(\theta\), the quantity \(m l^{2} \omega^{2} / 2\) is the kinetic energy of the pendulum while the term \(-m g l \cos \theta\) is the potential energy, referenced to the horizontal position \(\theta=\pi / 2\). The constant right-hand side is the total initial energy. (c) Determine the angular velocity at the instant the pendulum reaches the vertically downward position, \(\theta=0\). Express your answer in terms of the constants \(\omega_{0}\) and \(\theta_{0}\).

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ e^{t} y^{\prime}+(\cos y)^{2}=0, \quad y(0)=\pi / 4 $$

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ y \frac{d y}{d t}-\sin t=0, \quad y(\pi / 2)=-2 $$

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ t^{2} y^{\prime}+\sec y=0, \quad y(-1)=0 $$
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