Problem 82 - Calculus 3

Question

Find the unique solution satisfying the differential equation and the initial conditions given, where \(y_{p}(x)\) is the particular solution. \(y^{\prime \prime}+y=\cos x-4 \sin x\), \(y_{p}(x)=2 x \cos x+\frac{1}{2} x \sin x, \quad y(0)=8, \quad y^{\prime}(0)=-4\)

Step-by-Step Solution

Answer

The unique solution is \( y(x) = 8 \cos x + 2 \sin x + 2x \cos x + \frac{1}{2}x \sin x \).

Step 1: Identify the Differential Equation

The given differential equation is \( y'' + y = \cos x - 4 \sin x \). This is a non-homogeneous second-order differential equation, where we need to find the general solution.

Step 2: Particular Solution

The particular solution given is \( y_{p}(x) = 2x \cos x + \frac{1}{2}x \sin x \). This satisfies the non-homogeneous part of the differential equation.

Step 3: Find the Complementary Solution

The associated homogeneous equation is \( y'' + y = 0 \). The characteristic equation is \( r^2 + 1 = 0 \), which has roots \( r = \pm i \). Thus, the complementary solution is \( y_{c}(x) = C_1 \cos x + C_2 \sin x \).

Step 4: Write the General Solution

The general solution to the differential equation is the sum of the complementary and the particular solutions. Therefore, \( y(x) = y_{c}(x) + y_{p}(x) = C_1 \cos x + C_2 \sin x + 2x \cos x + \frac{1}{2}x \sin x \).

Step 5: Apply Initial Conditions

Using the initial conditions \( y(0) = 8 \) and \( y'(0) = -4 \), substitute \( x = 0 \) into the general solution to find the constants \( C_1 \) and \( C_2 \).- For \( y(0) = 8 \): Substitute into general solution: \[ C_1 \times \, \cos 0 + C_2 \times \, \sin 0 + 2 \times 0 \times \cos 0 + \frac{1}{2} \times 0 \times \sin 0 = 8 \]This simplifies to \( C_1 = 8 \).- For \( y'(0) = -4 \): First, find the derivative of the general solution. To do this, differentiate \( y(x) \) and apply \( x = 0 \) to find \( C_2 \). After differentiation, set:\[ -C_2 + 2 \cos 0 + 0 = -4 \] So, \( C_2 = 2 \).

Step 6: Formulate the Unique Solution

Substitute the values of \( C_1 = 8 \) and \( C_2 = 2 \) back into the general solution. Thus, the unique solution is:\[ y(x) = 8 \cos x + 2 \sin x + 2x \cos x + \frac{1}{2}x \sin x \]

Key Concepts

Particular Solution

A particular solution of a differential equation is a solution that includes no arbitrary constants and satisfies the entire differential equation including its non-homogeneous part. In the context of our problem, the non-homogeneous term is given by \( \cos x - 4 \sin x \).
The particular solution provided is \( y_{p}(x) = 2x \cos x + \frac{1}{2}x \sin x \). This is a specific formula tailored to cancel out or account for the non-homogeneous terms in the differential equation.
The importance of finding a particular solution lies in its ability to help solve the complete (general) solution when combined with a complementary solution. It sets a unique pathway for your answers in specific problems, letting you control for extra terms that standard methods may not handle. Particular solutions are vital in giving us the full context of real-world applications reflected in differential equations.

Complementary Solution

The complementary solution is derived from the homogeneous part of the differential equation, which ignores the non-homogeneous terms (in our case, \( \cos x - 4 \sin x \)). Essentially, it helps us understand the structure of solutions in the absence of external forces or inputs.
To find the complementary solution, you start by considering the homogeneous equation \( y'' + y = 0 \). This leads us to the characteristic equation \( r^2 + 1 = 0 \), where \( r \) are the roots of the equation. The solution here is \( r = \pm i \), pointing to a complementary solution of \( y_{c}(x) = C_1 \cos x + C_2 \sin x \).
What makes the complementary solution so crucial is that it outlines the behavior of the system's natural dynamics, which are then modified by the particular solution to accommodate external forces. Combining both solutions grants us a complete understanding of how systems respond both intrinsically and extrinsically.

Initial Conditions

Initial conditions are the specific values given for the solution of a differential equation at certain points. They are necessary to pinpoint a unique solution to a differential equation from a family of possible solutions. In our exercise, these are given as \( y(0) = 8 \) and \( y'(0) = -4 \).
Applying these conditions solves for the constants in our general solution, providing us with specific real-world parameters. Without these initial conditions, we could only determine a general solution with arbitrary constants \( C_1 \) and \( C_2 \) that don't reflect a specific scenario.
In practice, initial conditions are the initial states or prior-known solutions of a system, which is key in many fields such as physics and engineering, where predicting future behaviors based on known parameters and states is critical for planning and operations.