Problem 3 - Advanced Engineering Mathematics

Question

Solve each differential equation by variation of parameters. $$ y^{\prime \prime}+y=\sin x $$

Step-by-Step Solution

Answer

The general solution is: \(y = C_1 \cos x + C_2 \sin x - \frac{1}{2}(x + \sin x \cos x) \cos x + \frac{1}{2} \sin^2 x \sin x.\)

Step 1: Solve the Homogeneous Equation

First, solve the associated homogeneous equation: \(y^{\prime \prime} + y = 0\). The characteristic equation is: \(r^2 + 1 = 0\). Solving this, we get roots \(r = \pm i\). Hence, the general solution of the homogeneous equation is \(y_h = C_1 \cos x + C_2 \sin x\).

Step 2: Setup Particular Solution Using Variation of Parameters

For variation of parameters, we write the particular solution as \(y_p = u_1(x) \cos x + u_2(x) \sin x\), where \(u_1(x)\) and \(u_2(x)\) are functions to be determined.

Step 3: Compute Wronskian and Auxiliary Functions

Compute the Wronskian of the solutions \(\cos x\) and \(\sin x\): \[ W = \left| \begin{array}{cc} \cos x & \sin x \ -\sin x & \cos x \end{array} \right| = 1. \]Next, determine derivative of \(u_1\) and \(u_2\) using:\( u'_1(x) = -\frac{\sin x \cdot \sin x}{W} = -\sin^2 x, \) \( u'_2(x) = \frac{\cos x \cdot \sin x}{W} = \sin x \cdot \cos x.\)

Step 4: Integrate to Find u1 and u2

Integrate the expressions from Step 3:\( u_1(x) = \int -\sin^2 x \, dx = -\frac{1}{2}(x + \sin x \cos x), \)\( u_2(x) = \int \sin x \cos x \, dx = \frac{1}{2}\sin^2 x.\)

Step 5: Write the Particular Solution

Substitute \(u_1\) and \(u_2\) into the expression for \(y_p\):\( y_p = \left(-\frac{1}{2}(x + \sin x \cos x)\right) \cos x + \left(\frac{1}{2}\sin^2 x\right) \sin x.\)

Step 6: Combine Homogeneous and Particular Solutions

The general solution of the differential equation is the sum of the homogeneous and particular solutions:\[ y = C_1 \cos x + C_2 \sin x - \frac{1}{2}(x + \sin x \cos x) \cos x + \frac{1}{2} \sin^2 x \sin x. \]

Key Concepts

Homogeneous Differential Equation

A homogeneous differential equation is a type of differential equation where the terms are a function of the dependent variable and its derivatives only. In this context, "homogeneous" means that all terms are related to the dependent variable without any free-standing constants or particular external functions.
For example, in the differential equation \( y'' + y = 0 \), we have a perfect representation of a homogeneous equation. Here, both terms are dependent on the function \( y \) and its derivative.
Homogeneous equations often appear as a preliminary step when solving non-homogeneous equations like \( y'' + y = \sin x \). By first solving the homogeneous equation, we find the complementary function or general solution formed by the characteristic roots.
The solution to a homogeneous equation typically involves solving its characteristic equation. This leads us conveniently to our next concept.

Characteristic Equation

To find the solutions to homogeneous differential equations, we utilize the characteristic equation, an algebraic equation derived from the differential equation's highest-order derivative.
In the case of \( y'' + y = 0 \), forming the characteristic equation involves transforming the differential equation into a polynomial by replacing each derivative with a power of \( r \), like so: \( r^2 + 1 = 0 \).
Solving this equation for \( r \), we obtain the roots \( r = \pm i \). These roots are complex, which implies that the general solution of the homogeneous equation involves trigonometric functions. The solution becomes a combination of sines and cosines: \( y_h = C_1 \cos x + C_2 \sin x \).
These trigonometric functions form the basis solutions used in deriving the full general solution of a differential equation.

Wronskian

The Wronskian is a crucial concept when using methods like variation of parameters to solve differential equations. It helps determine the independence of solutions.
For two functions \( f(x) \) and \( g(x) \), the Wronskian \( W \) is calculated as follows:

• \( W = \begin{vmatrix} f & g \ f' & g' \end{vmatrix} \)

In our example, the solutions \( \cos x \) and \( \sin x \) are used. Their Wronskian is | \( \begin{vmatrix} \cos x & \sin x \ -\sin x & \cos x \end{vmatrix} \)| which equals 1. Since the Wronskian is non-zero, the solutions are linearly independent.
Linearly independent solutions mean they can be used to construct the general solution, as they'll span the solution space of the differential equation effectively.

General Solution

The general solution of a differential equation is a combination of the homogeneous solution and a particular solution that satisfies the entire equation. By adding these together, we capture all potential solutions to the equation.
For instance, after solving the homogeneous differential equation to obtain \( y_h = C_1 \cos x + C_2 \sin x \), we find the particular solution \( y_p \) using methods like variation of parameters.
In the example discussed, the particular solution is derived through integration, resulting in: \( y_p = -\frac{1}{2}(x + \sin x \cos x) \cos x + \frac{1}{2} \sin^2 x \sin x \).
Combining both the homogeneous and the particular solutions, we obtain the general solution:

• \( y = C_1 \cos x + C_2 \sin x - \frac{1}{2}(x + \sin x \cos x) \cos x + \frac{1}{2} \sin^2 x \sin x \)

This combined expression fully describes all behaviours of the differential equation's solutions.