Problem 32 - Advanced Engineering Mathematics

Question

Solve the given initial-value problem. \(y^{\prime \prime}-y=\cosh x, y(0)=2, y^{\prime}(0)=12\)

Step-by-Step Solution

Answer

The solution is \( y = 6.75e^x - 4.75e^{-x} + \frac{1}{4}xe^x + \frac{1}{4}xe^{-x} \).

Step 1: Solve the Homogeneous Equation

First, solve the homogeneous equation corresponding to the differential equation, which is: \[ y'' - y = 0 \]The characteristic equation is \( r^2 - 1 = 0 \), which factors to \( (r-1)(r+1) = 0 \) with roots \( r = 1, -1 \). Hence, the general solution of the homogeneous equation is: \[ y_h = C_1 e^x + C_2 e^{-x} \]

Step 2: Find a Particular Solution

Next, find a particular solution \( y_p \) to the non-homogeneous equation:\[ y'' - y = \cosh x \]Since \( \cosh x = \frac{e^x + e^{-x}}{2} \), we can try a particular solution of the form:\[ y_p = Ax e^x + Bxe^{-x} \]Plugging this form into the equation and assuming symmetry for simplicity:Solve using undetermined coefficients to find suitable values of A and B.

Step 3: Solve for Undetermined Coefficients

Compute the first and second derivatives of our assumed particular solution:\[ y_p' = Ax e^x + Ae^x + Bx e^{-x} + Be^{-x} \]\[ y_p'' = Ax e^x + 2Ae^x + Bx e^{-x} + 2Be^{-x} \]Substitute \( y_p, y_p' \) and \( y_p'' \) back into the non-homogeneous equation \( y'' - y = \cosh x \) to find:\[ 2Ae^x + 2Be^{-x} = \frac{e^x + e^{-x}}{2} \]This yields the equations \( 2A = \frac{1}{2} \) and \( 2B = \frac{1}{2} \), giving \( A = \frac{1}{4} \) and \( B = \frac{1}{4} \). So, \( y_p = \frac{1}{4}xe^x + \frac{1}{4}xe^{-x} \).

Step 4: Write the General Solution

The general solution to the initial value problem is the sum of the homogeneous and particular solutions:\[ y = C_1 e^x + C_2 e^{-x} + \frac{1}{4}xe^x + \frac{1}{4} xe^{-x} \]

Step 5: Apply Initial Conditions

Use the initial conditions \( y(0) = 2 \) and \( y'(0) = 12 \) to solve for \( C_1 \) and \( C_2 \). **First condition:**\[ y(0) = C_1 + C_2 = 2 \]**Second condition:** Find \( y'(x) = C_1 e^x - C_2 e^{-x} + \frac{1}{4}e^x + \frac{1}{4} e^{-x} \)Plug in initial condition:\[ y'(0) = C_1 - C_2 + \frac{1}{4} + \frac{1}{4} = 12 \]Solving gives:\[ C_1 + C_2 = 2 \]\[ C_1 - C_2 = 11.5 \]Solve simultaneously to find \( C_1 = 6.75 \) and \( C_2 = -4.75 \).

Step 6: Final Solution

Plugging \( C_1 \) and \( C_2 \) back into the general solution gives:\[ y = 6.75e^x - 4.75e^{-x} + \frac{1}{4}xe^x + \frac{1}{4}xe^{-x} \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. In other words, they express relationships between a function and its derivatives. These equations are central to fields like physics, engineering, and economics, as they model how systems change. For any given problem, solving a differential equation means finding the function or functions that satisfy the relationship described. In our exercise, the equation is of second order, meaning it involves the second derivative of an unknown function:

• The equation is given as: \( y'' - y = \cosh x \), where \( y'' \) is the second derivative of \( y \) with respect to \( x \).
• The problem involves initial conditions: \( y(0) = 2 \) and \( y'(0) = 12 \).
• These initial conditions help determine the specific solution that fits the problem scenario.

Differential equations can be homogeneous when all terms are dependent on the function or its derivatives, or non-homogeneous when there is an additional function (in our case, \( \cosh x \)) that does not depend on \( y \) or its derivatives.

Characteristic Equation

A characteristic equation is a tool used to solve linear differential equations with constant coefficients. It derives directly from the homogeneous part of a differential equation by assuming solutions of the form \( y = e^{rx} \). From this assumption, the derivatives \( y' \) and \( y'' \) turn into multiples of \( e^{rx} \), and substituting these into the original differential equation leads to an algebraic equation where \( r \) is the variable. Here's how this process works in our problem:

• Start with the homogeneous part of the equation: \( y'' - y = 0 \).
• Assume a solution of the form \( y = e^{rx} \).
• The characteristic equation becomes \( r^2 - 1 = 0 \).
• Solve this algebraic equation to find the roots \( r = 1 \) and \( r = -1 \).
• These roots indicate that the general solution for the homogeneous equation is \( y_h = C_1 e^x + C_2 e^{-x} \).

The roots of the characteristic equation help identify the possible types of solutions for the differential equation. Distinct real roots, as seen here, indicate exponential solutions.

Undetermined Coefficients

The method of undetermined coefficients is a technique used to find particular solutions to linear non-homogeneous differential equations. This method is useful when the non-homogeneous term (right-hand side of the equation) is simple, like polynomials, exponentials, or trigonometric functions. It involves:

• Guessing a form for the particular solution \( y_p \).
• The form is based on the non-homogeneous term, which is \( \cosh x \), expressed as \( \frac{e^x + e^{-x}}{2} \).
• We assume \( y_p = Ax e^x + Bxe^{-x} \).
• Then, calculate \( y_p' \) and \( y_p'' \) from this assumed form.
• Substitute \( y_p, y_p' \, \text{and} \, y_p'' \) back into the equation to equate coefficients and find the values of \( A \) and \( B \).
• This gives us \( A = B = \frac{1}{4} \).
• The particular solution then becomes \( y_p = \frac{1}{4}xe^x + \frac{1}{4}xe^{-x} \).

This procedure allows the conditions of the problem to determine a particular solution that fits within the context offered by the non-homogeneous term.

Non-Homogeneous Equation

A non-homogeneous equation is a differential equation that has a source or forcing function, making it different from its homogeneous counterpart. For these equations, the solution is a combination of two parts:

• The homogeneous solution (\( y_h \)), found by solving the equation when the non-homogeneous term is zero.
• The particular solution (\( y_p \)), which specifically accounts for the non-homogeneous component in the equation.

The general solution is then written as their sum: \( y = y_h + y_p \).For the equation \( y'' - y = \cosh x \):

• We first solved the homogeneous equation from the characteristic equation to get the general solution \( y_h = C_1 e^x + C_2 e^{-x} \).
• Then, using the method of undetermined coefficients, we derived a particular solution \( y_p = \frac{1}{4}xe^x + \frac{1}{4}xe^{-x} \).
• Combining these, the general solution becomes: \( y = C_1 e^x + C_2 e^{-x} + \frac{1}{4}xe^x + \frac{1}{4}xe^{-x} \).
• Finally, we use the initial conditions, \( y(0) = 2 \) and \( y'(0) = 12 \), to solve for \( C_1 \) and \( C_2 \), thus completing the initial value problem.

The solution encapsulates both the natural behavior of the system and the influence of external forces captured in the non-homogeneous term.