Problem 76 - Calculus Volume 3

Question

Solve the differential equation using the method of variation of parameters. $$ 4 y^{\prime \prime}+y=2 \sin x $$

Step-by-Step Solution

Answer

The general solution is \( y = C_1\cos\left(\frac{x}{2}\right) + C_2\sin\left(\frac{x}{2}\right) + y_p \) with \( y_p \) calculated in Step 6.

Step 1: Write down the given differential equation

The differential equation given in the problem is \( 4y'' + y = 2\sin x \). It is a second-order linear differential equation with constant coefficients.

Step 2: Find the complementary solution

First, solve the homogeneous equation: \( 4y'' + y = 0 \). Assume a solution of the form \( y = e^{rx} \), substituting this into the homogeneous equation yields the characteristic equation: \( 4r^2 + 1 = 0 \). Solving for \( r \), we get \( r = \pm \frac{i}{2} \), which gives us the complementary solution: \( y_c = C_1\cos\left(\frac{x}{2}\right) + C_2\sin\left(\frac{x}{2}\right) \).

Step 3: Setup for variation of parameters

When using variation of parameters, express the particular solution as \( y_p = u_1(x)\cos\left(\frac{x}{2}\right) + u_2(x)\sin\left(\frac{x}{2}\right) \). The functions \( u_1(x) \) and \( u_2(x) \) must satisfy the system: \( u_1'(x)\cos\left(\frac{x}{2}\right) + u_2'(x)\sin\left(\frac{x}{2}\right) = 0 \) and \( -u_1'(x)\sin\left(\frac{x}{2}\right) + u_2'(x)\cos\left(\frac{x}{2}\right) = \frac{2\sin x}{4} = \frac{\sin x}{2} \).

Step 4: Solve the system of equations for u_1'(x) and u_2'(x)

The Wronskian \( W = \begin{vmatrix} \cos\left(\frac{x}{2}\right) & \sin\left(\frac{x}{2}\right) \ -\frac{1}{2}\sin\left(\frac{x}{2}\right) & \frac{1}{2}\cos\left(\frac{x}{2}\right) \end{vmatrix} = \frac{1}{2} \). Then, \( u_1'(x) = \frac{-\sin\left(\frac{x}{2}\right) \cdot \frac{\sin x}{2}}{\frac{1}{2}} = -\sin\left(\frac{x}{2}\right)\sin x \) and \( u_2'(x) = \frac{\cos\left(\frac{x}{2}\right) \cdot \frac{\sin x}{2}}{\frac{1}{2}} = \cos\left(\frac{x}{2}\right)\sin x \).

Step 5: Integrate to find u_1(x) and u_2(x)

To find \( u_1(x) \) and \( u_2(x) \), integrate the above expressions: \( u_1(x) = \int -\sin\left(\frac{x}{2}\right)\sin x \, dx \) which simplifies to: \( u_1(x) = -2\cos x + x\sin x + C_3 \).Similarly, \( u_2(x) = \int \cos\left(\frac{x}{2}\right)\sin x \, dx \) simplifies to: \( u_2(x) = 2\cos^2\left(\frac{x}{2}\right) - C_4 \).

Step 6: Write the particular solution

Now substitute \( u_1(x) \) and \( u_2(x) \) into the particular solution \( y_p = u_1(x)\cos\left(\frac{x}{2}\right) + u_2(x)\sin\left(\frac{x}{2}\right) \). This gives the particular solution: \( y_p = (-2\cos x + x\sin x)\cos\left(\frac{x}{2}\right) + 2\cos^2\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) \).

Step 7: Write the general solution

Finally, the general solution to the differential equation is the sum of the complementary solution and the particular solution: \( y = C_1\cos\left(\frac{x}{2}\right) + C_2\sin\left(\frac{x}{2}\right) + (-2\cos x + x\sin x)\cos\left(\frac{x}{2}\right) + 2\cos^2\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They express a relationship between an unknown function and its derivatives.
Differential equations are essential in modeling various physical phenomena, such as motion, heat, and soundwaves.
Understanding differential equations is crucial for solving a wide range of real-world problems.

• They can be classified based on order: first-order, second-order, etc.
• Linear or non-linear: Linear equations have proportional relationships, while non-linear equations do not.

In the given exercise, we are dealing with a second-order linear differential equation, where the focus is on solutions that satisfy the equation.

Second-Order Linear Differential Equations

A second-order differential equation involves the second derivative of a function. In mathematical notation, this is usually represented as \( y''\).
These equations often arise in systems with acceleration or curvature, such as in mechanical and electrical systems.

• Second-order differential equations have the general form \( a y'' + b y' + c y = g(x)\).
• They can be homogeneous (where \(g(x) = 0\)) or non-homogeneous (where \(g(x) eq 0\)).

The exercise provided is a non-homogeneous equation because it includes the term \(2\sin x\). In such cases, the general solution is the sum of the complementary solution (related to the homogeneous equation) and a particular solution that satisfies the non-homogeneous part.

Complementary and Particular Solutions

The solution to a second-order linear differential equation with constant coefficients involves finding both complementary and particular solutions.
The complementary solution, \( y_c \), is related to the homogeneous equation \( 4y'' + y = 0 \).
By solving this, we find \( y_c = C_1\cos\left(\frac{x}{2}\right) + C_2\sin\left(\frac{x}{2}\right) \).

• The complementary solution solves the homogeneous part of the differential equation.
• It involves constants \( C_1 \) and \( C_2 \), which are determined by initial conditions.

For the particular solution, the method of variation of parameters allows us to incorporate the non-homogeneous elements (like \(2\sin x\)) into the solution. This requires setting up a system of equations for \( u_1(x) \) and \( u_2(x) \) that ensure the resulting solution satisfies the original differential equation.
Together, the complementary and the particular solutions form the general solution. For the given equation, the general solution is:\[ y = C_1\cos\left(\frac{x}{2}\right) + C_2\sin\left(\frac{x}{2}\right) + (-2\cos x + x\sin x)\cos\left(\frac{x}{2}\right) + 2\cos^2\left(\frac{x}{2}\right)\sin\left(\frac{x}{2}\right) \]