Question

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

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

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 \(x\).

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.

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.

Combine Solutions

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

Key concepts

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 \( E \).