Problem 37 - Advanced Engineering Mathematics

Question

In Problems 37 and 38, solve the given initial-value problem. $$ \frac{d^{2} x}{d t^{2}}+4 x=-5 \sin 2 t+3 \cos 2 t, x(0)=-1, x^{\prime}(0)=1 $$

Step-by-Step Solution

Answer

The solution to the initial-value problem is \( x(t) = -\frac{1}{2} \cos 2t + \frac{7}{2} \sin 2t \).

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{d^{2}x}{dt^{2}} + 4x = -5 \sin 2t + 3 \cos 2t \). This is a second-order linear non-homogeneous differential equation.

Step 2: Solve the Homogeneous Equation

The corresponding homogeneous equation is \( \frac{d^{2}x}{dt^{2}} + 4x = 0 \). Solving this gives the characteristic equation \( r^2 + 4 = 0 \). Hence, \( r = \pm 2i \), leading to the complementary solution \( x_c(t) = C_1 \cos 2t + C_2 \sin 2t \).

Step 3: Solve the Particular Equation

We use the method of undetermined coefficients to find a particular solution for the non-homogeneous part. Assume a particular solution of the form \( x_p(t) = A \sin 2t + B \cos 2t \). Substitute into the original equation, differentiate as needed, and equate coefficients to solve for \( A \) and \( B \).

Step 4: Substitute and Solve for Coefficients

Substituting \( x_p(t) = A \sin 2t + B \cos 2t \) into \( \frac{d^{2}x}{dt^{2}} + 4x = -5 \sin 2t + 3 \cos 2t \) gives \( -4A \sin 2t - 4B \cos 2t + 4A \sin 2t + 4B \cos 2t = -5 \sin 2t + 3 \cos 2t \). Solving for coefficients yields \( A = \frac{5}{4} \), \( B = \frac{3}{4} \).

Step 5: Form the General Solution

The general solution is the sum of the complementary and particular solutions: \( x(t) = C_1 \cos 2t + C_2 \sin 2t + \frac{5}{4} \sin 2t + \frac{3}{4} \cos 2t \).

Step 6: Apply the Initial Conditions

Use the initial conditions \( x(0) = -1 \) and \( x'(0) = 1 \) to solve for \( C_1 \) and \( C_2 \). For \( x(0) = -1 \), we have \( C_1 + \frac{3}{4} = -1 \), so \( C_1 = -\frac{7}{4} \). For \( x'(0) = 1 \), differentiate \( x(t) \) and substitute to solve for \( C_2 \). This gives \( C_2 = \frac{9}{4} \).

Step 7: Write the Final Solution

Substituting \( C_1 = -\frac{7}{4} \), \( C_2 = \frac{9}{4} \) into the general solution, we get: \[ x(t) = -\frac{7}{4} \cos 2t + \frac{9}{4} \sin 2t + \frac{5}{4} \sin 2t + \frac{3}{4} \cos 2t \] Simplify to get the final solution: \[ x(t) = -\frac{1}{2} \cos 2t + \frac{7}{2} \sin 2t \].

Key Concepts

Homogeneous Equations

In mathematics, homogeneous equations are a type of differential equation where the terms involving the dependent variable and its derivatives are directly set to zero. This means these equations have a form something like \( \frac{d^2x}{dt^2} + 4x = 0 \). Here, the left side involves derivatives of \( x \) and \( x \) itself, but all terms sum to zero.
This equation is significant because it helps us find the complementary solution, which is a core component of solving non-homogeneous differential equations.

The importance lies in establishing a basis for the solution. By solving homogeneous equations, you find solutions that form the basis for the solutions to more complex non-homogeneous equations. The characteristic equation, discussed below, is used to find these solutions.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find a particular solution to non-homogeneous linear differential equations. When an equation involves terms that aren't zero on the right side, as in \( \frac{d^2x}{dt^2} + 4x = -5 \sin 2t + 3 \cos 2t \), we use this method.

Here's how it works:

• Assume a particular form for the solution based on the non-homogeneous part. Here, we assume \( x_p(t) = A \sin 2t + B \cos 2t \).
• Substitute this assumed form into the equation.
• Differentiate as necessary and equate coefficients to solve for unknowns, aiming to find the values of \( A \) and \( B \).

This approach is powerful because it allows mathematicians to solve equations that appear complex by breaking them down into solvable components. Once we find \( A \) and \( B \), we can write an entire particular solution.

Initial-Value Problem

An initial-value problem (IVP) is a differential equation paired with specific conditions known as 'initial conditions.' These allow us to pinpoint a unique solution for the differential equation. In our problem, the initial conditions are \( x(0) = -1 \) and \( x'(0) = 1 \).

These conditions describe the state of a system at the starting point—often, time \( t = 0 \). They guide us in finding the specific constants in our general solution, which makes the solution unique rather than broadly applicable.

• Initial conditions provide specific values for the function and its derivatives at a certain point.
• They are crucial because they help tailor a general solution to fit particular scenarios.

By applying these conditions, we find the constants in homogeneous or general solutions that satisfy the overall system within the given constraints.

Characteristic Equation

The characteristic equation is a crucial part of solving homogeneous differential equations. It's derived from the differential equation itself and helps find the complementary solution. For the equation \( \frac{d^2x}{dt^2} + 4x = 0 \), the characteristic equation is formed by replacing derivatives with powers of \( r \), giving us \( r^2 + 4 = 0 \).

Solving this equation determines the values of \( r \), which represent the eigenvalues.

• These values, like \( r = \pm 2i \) in our example, explain the type of solutions: real, complex, or repeated roots.
• These solutions lead to discovering the complementary solution \( C_1 \cos 2t + C_2 \sin 2t \).

The characteristic equation thus bridges the original differential equation and its solutions, unlocking pathways to understanding the dynamics described by the equation.