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Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime}-2 y^{\prime}+5 y=5 x+4 e^{x}(1+\sin 2 x)$$

Short Answer

Expert verified
The differential equation is a second-order linear non-homogeneous ODE. The solution combines the homogeneous solution \(y_h = e^x(c_1 \cos(2x) + c_2 \sin(2x))\) and a particular solution \(y_p\).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \(y'' - 2y' + 5y = 5x + 4e^x(1 + \sin 2x)\). This is a second-order linear non-homogeneous ordinary differential equation because the highest derivative is second-order, it has the form \(ay'' + by' + cy = g(x)\), and \(g(x)\) is a non-zero function of \(x\).
02

Solve the Corresponding Homogeneous Equation

First, solve the homogeneous equation \(y'' - 2y' + 5y = 0\). The characteristic equation is \( r^2 - 2r + 5 = 0 \). Solving this quadratic equation, \[r = \frac{2 \pm \sqrt{4 - 20}}{2} = 1 \pm 2i\]. Therefore, the general solution of the homogeneous equation is \(y_h = e^x(c_1 \cos(2x) + c_2 \sin(2x))\), where \(c_1\) and \(c_2\) are arbitrary constants.
03

Find a Particular Solution

To find a particular solution \(y_p\), we use the method of undetermined coefficients. For the non-homogeneous part \(5x + 4e^x(1 + \sin 2x)\), consider the trial solution: \[y_p = Ax + B + e^x(C + Dx \sin 2x + Ex \cos 2x)\]. Substituting \(y_p\) into the original differential equation, determine coefficients \(A, B, C, D,\) and \(E\) by equating coefficients of like terms.
04

Combine Solutions

The general solution of the differential equation is the sum of the general solution of the homogeneous equation and the particular solution: \[y = y_h + y_p = e^x(c_1 \cos(2x) + c_2 \sin(2x)) + y_p\].

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Key Concepts

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

second-order differential equations
A second-order differential equation is an equation that involves the second derivative of a function. In mathematical terms, it can be written as: \[ a(x) y'' + b(x) y' + c(x) y = g(x) \]. Here, \( y \) is the unknown function and \( a(x), b(x), \) and \( c(x) \) are known functions, while \( g(x) \) can be zero or a non-zero function of \( x \). If \( g(x) \) is zero, the equation is called homogeneous. Otherwise, it is non-homogeneous.

Solving second-order differential equations often involves finding two solutions: the general solution to the homogeneous equation and a particular solution to the non-homogeneous one.

To solve these equations, we typically perform the following steps:
  • Identify the type of differential equation.
  • Solve the corresponding homogeneous equation.
  • Find a particular solution for the non-homogeneous equation.
  • Combine the two solutions to get the general solution.
linear non-homogeneous differential equations
A linear non-homogeneous differential equation has the general form: \[ a(x) y'' + b(x) y' + c(x) y = g(x) \].

The right-hand side (\( g(x) \)) is a non-zero function, making it non-homogeneous. Such equations represent physical systems where an external force or input is present.

For example, our given differential equation is \( y'' - 2y' + 5y = 5x + 4e^x(1+\text{sin}(2x)) \). Here, the function \( g(x) = 5x + 4e^x (1+\text{sin}(2x)) \) is non-zero, making the equation non-homogeneous.

Steps to solve include:
  • Solve the homogeneous part first (i.e., \( y'' - 2y' + 5y = 0 \)). This involves finding the characteristic equation (a quadratic in \( r \)) and solving for the roots.
  • For our example, the characteristic equation \( r^2 - 2r + 5 = 0 \) leads to roots \( r = 1 \pm 2i \), which provides our complementary solution \( y_h = e^x(c_1 \text{cos}(2x) + c_2 \text{sin}(2x)) \).
  • Identify a form for \( y_p \) (particular solution) that reflects the term \( g(x) \). Use the method of undetermined coefficients to find the specific form substituting back into the original equation.
method of undetermined coefficients
The method of undetermined coefficients is used to find a particular solution to a non-homogeneous linear differential equation. It works best when \( g(x) \) is a simple function, such as polynomials, exponentials, sines and cosines, or their finite sums/products.

The process involves:
  • Guessing the form of the particular solution based on \( g(x) \)'s form.
  • Substituting this guess into the original differential equation.
  • Solving for the unknown coefficients by equating terms on both sides of the equation.
For our example equation, \( y'' - 2y' + 5y = 5x + 4e^x(1+\text{sin}(2x)) \), the trial solution will be: \[ y_p = Ax + B + e^x(C + Dx \text{sin}(2x) + Ex \text{cos}(2x)) \].

Here we break it down:
  • The term \( 5x \) suggests a guess like \( Ax + B \).
  • The term \( 4e^x(1+\text{sin}(2x)) \) suggests a guess like \( e^x(C + Dx \text{sin}(2x) + Ex \text{cos}(2x)) \).
Using this combined form, substituting into the original equation will help us determine the coefficients \( A, B, C, D, \) and \( E \).

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Most popular questions from this chapter

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$\frac{d^{2} r}{d t^{2}}-6 \frac{d r}{d t}+9 r=0$$

Compare the temperatures of your cup of coffee at time \(t\) (a) if you add cream and let the mixture cool; (b) if you let the coffee and cream sit on the table and mix them at time \(t\). Hints: Assume Newton's law of cooling (Problem 2.27) for both coffee and cream (where it is a law of heating). Combine \(n^{\prime}\) units of cream initially at temperature \(T_{0}^{\prime}\) with \(n\) units of coffee initially at temperature \(T_{0},\) and find the temperature at time \(t\) in (a) and in (b) assuming that the air temperature remains a constant \(T_{a},\) and that the proportionality constant in the law of cooling is the same for both coffee and cream.

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}+y^{\prime}-5 y=e^{2 t}, \quad y_{0}=1, y_{0}^{\prime}=2$$

An object of mass \(m\) falls from rest under gravity subject to an air resistance proportional to its speed. Taking the \(y\) axis as positive down, show that the differential equation of motion is \(m(d v / d t)=m g-k v,\) where \(k\) is a positive constant. Find \(v\) as a function of \(t,\) and find the limiting value of \(v\) as \(t\) tends to infinity; this limit is called the terminal speed. Can you find the terminal speed directly from the differential equation without solving it? Hint: What is \(d v / d t\) after \(v\) has reached an essentially constant value? Consider the following specific examples of this problem. (a) A person drops from an airplane with a parachute. Find a reasonable value of \(k\) (b) In the Millikan oil drop experiment to measure the charge of an electron, tiny electrically charged drops of oil fall through air under gravity or rise under the combination of gravity and an electric field. Measurements can be made only after they have reached terminal speed. Find a formula for the time required for a drop starting at rest to reach 99\% of its terminal speed.

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$(2 x-y \sin 2 x) d x=\left(\sin ^{2} x-2 y\right) d y$$

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