Problem 1 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The general solution is \(y = C_1 \cos x + C_2 \sin x + 1\).

Step 1: Find the Complementary Solution

First, solve the homogeneous differential equation associated with the given problem. The homogeneous equation is \(y'' + y = 0\). This is a second-order linear differential equation with constant coefficients. The characteristic equation is \(r^2 + 1 = 0\), which has roots \(r = i\) and \(r = -i\). Therefore, the complementary solution is \(y_c = C_1 \cos x + C_2 \sin x\).

Step 2: Set Up Particular Solution Form

For the particular solution, we use the variation of parameters. The particular solution takes the form \(y_p = u_1(x) \cos x + u_2(x) \sin x\), where \(u_1(x)\) and \(u_2(x)\) are functions we need to determine. The goal is to find expressions for \(u_1\) and \(u_2\).

Step 3: Compute the Wronskian

The Wronskian, \(W\), of two functions \(\cos x\) and \(\sin x\) is computed as follows: \(W = \begin{vmatrix} \cos x & \sin x \ -\sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1\). The Wronskian is 1, which simplifies our calculations for \(u_1(x)\) and \(u_2(x)\).

Step 4: Determine Expressions for u1 and u2

Use the formulas for \(u_1'\) and \(u_2'\): \[ u_1' = -\frac{\sin x \cdot \sec x}{W} = -\sin x \] \[ u_2' = \frac{\cos x \cdot \sec x}{W} = \cos x \] Integrate these expressions to find \(u_1(x)\) and \(u_2(x)\):\[ u_1(x) = \int -\sin x \, dx = \cos x \] \[ u_2(x) = \int \cos x \, dx = \sin x \]

Step 5: Construct the Particular Solution

Substitute \(u_1(x)\) and \(u_2(x)\) back into the form for the particular solution: \(y_p = \cos x \cdot \cos x + \sin x \cdot \sin x = \cos^2 x + \sin^2 x\).Since \(\cos^2 x + \sin^2 x = 1\), this simplifies to \(y_p = 1\).

Step 6: Write the General Solution

Combine the complementary solution and the particular solution to get the general solution:\(y = C_1 \cos x + C_2 \sin x + 1\), where \(C_1\) and \(C_2\) are constants.

Key Concepts

Complementary Solution

The complementary solution is a pivotal element in solving linear differential equations. For a given differential equation, the complementary solution is the solution to its corresponding homogeneous equation. To find it, solve the differential equation where the right-hand side is zero. In our exercise, the homogeneous form is:
\( y'' + y = 0 \).This is where the complementary solution is derived. By solving this, we utilize the roots of its characteristic equation. The solution encapsulates components that are homogeneous in nature:

• \( y'' + y = 0 \) is the homogeneous differential equation.
• The characteristic equation formed is \( r^2 + 1 = 0 \). This emerges from the assumption \( y = e^{rx} \).
• This characteristic equation has roots \( r = i \) and \( r = -i \), indicating complex solutions.

These roots lead us to:
The complementary solution: \[ y_c = C_1 \cos x + C_2 \sin x \] Here, \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions.

Wronskian

The Wronskian is a determinant-based technique crucial in assessing the linear independence of solutions to differential equations.For two functions, the Wronskian helps verify whether they are linearly independent. In simpler terms, if the Wronskian is non-zero, the functions are indeed independent.
In our exercise involving \( \cos x \) and \( \sin x \), the Wronskian is computed as:\[ W = \begin{vmatrix} \cos x & \sin x \ -\sin x & \cos x \end{vmatrix} \]This results in:\[ W = \cos^2 x + \sin^2 x = 1 \]The result, \( W = 1 \), means that \( \cos x \) and \( \sin x \) are linearly independent.
This finding aids in simplifying calculations while using methods like the variation of parameters.

Characteristic Equation

The characteristic equation is an algebraic equation that emerges from a differential equation. It plays a crucial role in identifying the nature of solutions for differential equations.In our example, the differential equation:\[ y'' + y = 0 \]leads to the characteristic equation:\[ r^2 + 1 = 0 \]This equation typically arises by supposing a solution of the form \( y = e^{rx} \). Solving this characteristic equation gives the roots:

• \( r = i \)
• \( r = -i \)

Each root provides valuable insight into the solution's structure.
Since we obtain complex roots, it suggests the solution contains trigonometric functions, specifically \( \cos \) and \( \sin \), as observed in our complementary solution: \[ y_c = C_1 \cos x + C_2 \sin x \]Understanding the characteristic equation is essential as it guides us in formulating the complementary solution.

Particular Solution

The particular solution arises when solving non-homogeneous differential equations. It accounts for the non-zero component on the equation's right-hand side.For our exercise, the non-homogeneous equation is:\[ y'' + y = \sec x \]When using the variation of parameters, the particular solution is constructed by finding functions \( u_1(x) \) and \( u_2(x) \) that depend on the non-homogeneous part. The form for the particular solution is:\[ y_p = u_1(x) \cos x + u_2(x) \sin x \]The computation involves solving the integrals:

• \( u_1'(x) = -\sin x \)
• \( u_2'(x) = \cos x \)

Integrating, we get:

• \( u_1(x) = \cos x \)
• \( u_2(x) = \sin x \)

Substituting these back, the particular solution simplifies to:\[ y_p = \cos^2 x + \sin^2 x = 1 \]This simple result is due to the Pythagorean identity. Ultimately, the combination of both the complementary and particular solutions form the full general solution to the differential equation.