Problem 14 - Advanced Engineering Mathematics

Question

In Problems \(1-18\), solve each differential equation by variation of parameters. $$ y^{\prime \prime}-2 y^{\prime}+y=e^{t} \arctan t $$

Step-by-Step Solution

Answer

Solve the homogeneous equation, propose a particular solution, and use variation of parameters to find the particular integral. The full solution is the sum of the homogeneous and particular solutions.

Step 1: Identify the Homogeneous Equation

The given differential equation is \( y'' - 2y' + y = e^{t} \arctan(t) \). First, we need to find the corresponding homogeneous equation, which is \( y'' - 2y' + y = 0 \).

Step 2: Solve the Homogeneous Equation

The characteristic equation of the homogeneous equation \( y'' - 2y' + y = 0 \) is \( r^2 - 2r + 1 = 0 \). This is a perfect square, \( (r - 1)^2 = 0 \), which means \( r = 1 \) is a double root. Hence, the solution to the homogeneous equation is \( y_h = C_1 e^t + C_2 t e^t \).

Step 3: Set Up the Variation of Parameters

Since the solution of the homogeneous equation is \( y_h = C_1 e^t + C_2 t e^t \), we use variation of parameters to find a particular solution. We suppose \( y_p = u_1 e^t + u_2 t e^t \), where \( u_1 \) and \( u_2 \) are functions to be determined.

Step 4: Differentiate the Proposed Solution

Calculate the derivatives: \( y_p' = (u_1' e^t + u_2' t e^t) + (u_1 + u_2 t) e^t \) and \( y_p'' = (u_1' e^t + u_2' t e^t + u_2 e^t) + 2(u_1 + u_2 t) e^t \).

Step 5: Apply Variation of Parameters Conditions

The conditions for variation of parameters are \( u_1' e^t + u_2' t e^t = 0 \) and \( u_1' e^t + u_2' (t e^t + e^t) = e^t \arctan(t) \).

Step 6: Solve the System of Equations

From the first condition, solve \( u_1' e^t + u_2' t e^t = 0 \) to get \( u_1' = -u_2' t \). Substitute this into the second condition to get \( -u_2' t e^t + u_2' (t e^t + e^t) = e^t \arctan(t) \). Simplify to find \( u_2' e^t = e^t \arctan(t) \), hence \( u_2' = \arctan(t) \).

Step 7: Integrate to Find \( u_2 \)

Integrate \( u_2' = \arctan(t) \) to find \( u_2 \). Use the integration by parts or look up the table of integrals to get \( u_2 = t \arctan(t) - \frac{1}{2}\ln(1+t^2) + C \).

Step 8: Find \( u_1 \) from \( u_2 \)

Use \( u_1' = -u_2' t \) to solve for \( u_1 \). First, differentiate: \( u_2' = \arctan(t) \). Since \( u_1' = -t \arctan(t) \), integrate to get \( u_1(t) = -\int t \arctan(t) dt \). Calculating this integral by parts, the result is \( u_1 = -\frac{t^2}{2}\arctan(t) + \frac{1}{4}t^2 \ln(1+t^2) - \frac{1}{8}(t^2 + 2) + D \).

Step 9: Compile the Particular Solution

Substitute \( u_1 \) and \( u_2 \) back into \( y_p = u_1 e^t + u_2 t e^t \) to find the particular solution. Compute and simplify to express in terms of components related to \( e^t \).

Step 10: Write the General Solution

The general solution to the differential equation is the sum of the homogeneous solution and the particular solution: \[ y = C_1 e^t + C_2 t e^t + y_p \]. Substitute the expressions found for \( C_1 \, C_2 rom \, and \, y_p \) to obtain the full solution.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are essential in modeling situations where changes occur continuously or dynamically over time, such as in physics, engineering, and biology. In our exercise, we focus on a second-order linear differential equation, which involves the second derivative of the unknown function. A second-order linear equation has the general form:

• \( a y'' + b y' + c y = g(t) \)

where \( a \), \( b \), and \( c \) are constants or functions of the independent variable \( t \), and \( g(t) \) is known as the non-homogeneous term. Solving these equations typically involves finding two parts of the solution: the homogeneous solution and the particular solution, combined to provide the general solution.

Homogeneous Solution

The homogeneous solution of a differential equation refers to solving a related equation where the non-homogeneous term is zero. For our given differential equation \( y'' - 2y' + y = e^t \arctan(t) \), the corresponding homogeneous equation is \( y'' - 2y' + y = 0 \).
The first step in solving the homogeneous equation is to find the characteristic equation:

• \( r^2 - 2r + 1 = 0 \)

This solves to \( (r-1)^2 = 0 \), indicating a double root \( r = 1 \). The double root reflects that the homogeneous solution will be composed of terms involving \( e^t \):

• \( y_h = C_1 e^t + C_2 te^t \)

where \( C_1 \) and \( C_2 \) are constants determined by initial conditions if provided.

Particular Solution

The particular solution caters to the effects of the non-homogeneous term \( g(t) \) in the differential equation. We use a method called variation of parameters to determine a function that satisfies this part. Our assumption for the particular solution is expressed as:

• \( y_p = u_1 e^t + u_2 t e^t \)

where \( u_1 \) and \( u_2 \) are functions of \( t \) that we need to determine. This involves substituting \( y_p \) and its derivatives back into the original equation and applying specific conditions. These conditions help us set up a system of equations from which \( u_1 \) and \( u_2 \) are derived by integration. Ultimately, these solutions alongside their derivatives reconstruct our particular solution tailored to the original non-homogeneous equation.

Characteristic Equation

The characteristic equation is crucial in finding the homogeneous solution of a linear differential equation. It is derived by assuming a solution of the form \( y = e^{rt} \), where \( r \) is a constant. Substituting this into the homogeneous equation leads to the characteristic equation.

• For the differential equation \( y'' - 2y' + y = 0 \), the characteristic equation is \( r^2 - 2r + 1 = 0 \).

Solving this equation gives the roots of the characteristic polynomial, which determine the form of the homogeneous solution. In this particular case, the equation is a perfect square, \( (r-1)^2 = 0 \), indicating a repeated root \( r = 1 \). This results in a solution structure of \( y_h = C_1 e^t + C_2 te^t \), where the repeated root introduces the extra factor of \( t \) multiplying one of the exponential terms. This step is fundamental in outlining the blueprint for the general solution.