Problem 29 - Advanced Engineering Mathematics

Question

Write down the form of the general solution \(y=y_{c}+y_{p}\) of the given differential equation in the two cases \(\omega \neq \alpha\) and \(\omega=\alpha\). Do not determine the coefficients in \(y_{p}\). (a) \(y^{\prime \prime}+\omega^{2} y=\sin \alpha x\) (b) \(y^{\prime \prime}-\omega^{2} y=e^{\alpha x}\)

Step-by-Step Solution

Answer

(a) Case 𝜔≠𝛼: \(y = C_1 \cos(\omega x) + C_2 \sin(\omega x) + A \cos(\alpha x) + B \sin(\alpha x)\); Case 𝜔=𝛼: \(y = C_1 \cos(\omega x) + C_2 \sin(\omega x) + Ax \cos(\alpha x) + Bx \sin(\alpha x)\). (b) Case 𝜔≠𝛼: \(y = C_1 e^{\omega x} + C_2 e^{-\omega x} + Ae^{\alpha x}\); Case 𝜔=𝛼: \(y = C_1 e^{\omega x} + C_2 e^{-\omega x} + Axe^{\alpha x}\).

Step 1: Solve the Homogeneous Equation (a)

The homogeneous equation is given by \( y'' + \omega^2 y = 0 \). This is a second-order linear differential equation with constant coefficients. The characteristic equation is \( r^2 + \omega^2 = 0 \), which gives roots \( r = \pm i\omega \). Therefore, the complementary function (homogeneous solution) is \( y_c = C_1 \cos(\omega x) + C_2 \sin(\omega x) \).

Step 2: Particular Solution for Case (a) when 𝜔≠𝛼

When \( \omega eq \alpha \), the non-homogeneous term \( \sin(\alpha x) \) suggests using a particular solution of the form \( y_p = A \cos(\alpha x) + B \sin(\alpha x) \). However, we don't determine \( A \) and \( B \) here.

Step 3: Formula for General Solution (a) when 𝜔≠𝛼

Combine the complementary function and particular solution: \( y = y_c + y_p = C_1 \cos(\omega x) + C_2 \sin(\omega x) + A \cos(\alpha x) + B \sin(\alpha x) \).

Step 4: Particular Solution for Case (a) when 𝜔=𝛼

When \( \omega = \alpha \), a solution of the form \( A \cos(\alpha x) + B \sin(\alpha x) \) would overlap with the complementary solution. Instead, use \( y_p = Ax \cos(\alpha x) + Bx \sin(\alpha x) \) to ensure linear independence.

Step 5: Formula for General Solution (a) when 𝜔=𝛼

The general solution is given by \( y = y_c + y_p = C_1 \cos(\omega x) + C_2 \sin(\omega x) + Ax \cos(\alpha x) + Bx \sin(\alpha x) \).

Step 6: Solve the Homogeneous Equation (b)

The homogeneous equation for part (b) is \( y'' - \omega^2 y = 0 \). The characteristic equation is \( r^2 - \omega^2 = 0 \), with roots \( r = \pm \omega \). This yields a complementary function \( y_c = C_1 e^{\omega x} + C_2 e^{-\omega x} \).

Step 7: Particular Solution for (b)

The non-homogeneous term \( e^{\alpha x} \) implies a particular solution of the form \( y_p = Ae^{\alpha x} \) if \( \alpha eq \omega \). If \( \alpha = \omega \), try a solution of the form \( y_p = Axe^{\alpha x} \).

Step 8: Formula for General Solution (b) when 𝜔≠𝛼

For \( \omega eq \alpha \), the general solution is \( y = y_c + y_p = C_1 e^{\omega x} + C_2 e^{-\omega x} + Ae^{\alpha x} \).

Step 9: Formula for General Solution (b) when 𝜔=𝛼

If \( \omega = \alpha \), the general solution becomes \( y = y_c + y_p = C_1 e^{\omega x} + C_2 e^{-\omega x} + Axe^{\alpha x} \).

Key Concepts

Homogeneous Differential Equations

In the study of differential equations, homogeneous differential equations are a foundational concept. These equations have the general form where all terms involve the dependent variable or its derivatives. For a second-order linear differential equation with constant coefficients, the structure is usually:

• \(y'' + ext{something} imes y = 0\)

The goal is to find the general solution to this equation. This is known as the complementary solution, represented as \(y_c\).
To solve such an equation, you start by finding the characteristic equation, which arises from assuming a solution of the form \(y = e^{rx}\). This results in an algebraic equation.

• For example, given \(y'' + ext{something} imes y = 0\), the characteristic equation could be \(r^2 + ext{something} = 0\).

Solve this for \(r\) to find the roots.
The nature of the roots determines the form of \(y_c\):

• Real and distinct roots lead to exponential solutions like \(C_1 e^{r_1 x} + C_2 e^{r_2 x}\).
• Complex conjugate roots lead to solutions involving sines and cosines, like \(C_1 \cos(kx) + C_2 \sin(kx)\).

These solutions are vital for building up the complete solution to a differential equation.

Particular Solution

The particular solution, denoted by \(y_p\), is a solution that specifically addresses the non-homogeneous part of a differential equation. When you have a differential equation like \(y'' + ext{something} \times y = ext{non-homogeneous term}\), you're looking for a solution that will satisfy this entire equation, not just the homogeneous part.
To find \(y_p\), you'll use methods like undetermined coefficients or variation of parameters, each appropriate for different types of non-homogeneous terms:

• If the non-homogeneous term is a sine or cosine function, try a particular solution \(A \cos( ext{term}) + B \sin( ext{term})\).
• If it's an exponential function like \(e^{ ext{term} x}\), try \(A e^{ ext{term} x}\).

When the non-homogeneous term is similar to part of the complementary solution, you need to adjust \(y_p\) to maintain linear independence. You do this by introducing factors like \(x\) (e.g., \(x \sin( ext{term} x)\) or \(x e^{ ext{term} x}\)).
This clever adjustment helps ensure that the particular solution is neither overshadowed by nor redundant with the complementary solution.

General Solution

The general solution of a differential equation is the heart of the solution process and is given as \(y = y_c + y_p\). It combines:

• The complementary solution \(y_c\), which solves the homogeneous part.
• The particular solution \(y_p\), which solves the non-homogeneous part.

Together, they ensure the solution satisfies the full original equation. It's like piecing together two halves to solve the puzzle completely.
Whether you're dealing with \(\omega eq \alpha\) or \(\omega = \alpha\), integrating \(y_c\) with \(y_p\) addresses both the inherent characteristics of the system and any external influences present in the equation's expression.
Remember, the coefficients in \(y_c\) are often determined by initial or boundary conditions given in the problem, ensuring the general solution is tailored to specific scenarios. This means while \(y_c\) offers a broad range of solutions, the particular solution narrows it down to meet exact conditions set by the non-homogeneous part.