Problem 30 - Advanced Engineering Mathematics

Question

(a) Given that \(y=\sin x\) is a solution of \(y^{(4)}+2 y^{\prime \prime \prime}+11 y^{\prime \prime}+\) \(2 y^{\prime}+10 y=0\), find the general solution of the DE without the aid of a calculator or a computer. (b) Find a linear second-order differential equation with constant coefficients for which \(y_{1}=1\) and \(y_{2}=e^{-x}\) are solutions of the associated homogeneous equation and \(y_{p}=\frac{1}{2} x^{2}-x\) is a particular solution of the nonhomogeneous equation.

Step-by-Step Solution

Answer

(a) General solution: \( y = c_1 \sin x + c_2 \cos x + e^{-x}(c_3 \cos(3x) + c_4 \sin(3x)) \). (b) Differential equation: \( y'' + y' = \frac{1}{2}x^2 - x \).

Step 1: Verify the given solution satisfies the differential equation

You are given that \( y = \sin x \) is a solution for the differential equation \( y^{(4)} + 2y^{'''} + 11y^{''} + 2y' + 10y = 0 \). First calculate the derivatives of \( y = \sin x \) up to the fourth order: \( y' = \cos x \), \( y'' = -\sin x \), \( y''' = -\cos x \), and \( y^{(4)} = \sin x \). Substitute these into the differential equation to verify it equals zero:\( \sin x + 2(-\cos x) + 11(-\sin x) + 2(\cos x) + 10\sin x = 0 \).Combine terms to see that this simplifies to zero, confirming \( y = \sin x \) is a solution.

Step 2: Write the characteristic equation

The differential equation \( y^{(4)} + 2y^{'''} + 11y^{''} + 2y' + 10y = 0 \) is linear with constant coefficients. Assume a solution of the form \( y = e^{rx} \). Substitute into the differential equation, resulting in the characteristic equation:\( r^4 + 2r^3 + 11r^2 + 2r + 10 = 0 \).

Step 3: Solve the characteristic equation

Use polynomial factorization and the given root (\( r = i \)) to factor the polynomial. If \( y = \sin x \) is a solution, \( y = \cos x \) must also be a solution. Therefore, the terms \( r = i \) and \( r = -i \) must be roots. We can factor the characteristic equation as \((r^2 + 1)\) and then use polynomial division to find additional roots:1. Divide \( r^4 + 2r^3 + 11r^2 + 2r + 10 \) by \( r^2 + 1 \).2. The quotient is \( r^2 + 2r + 10 \), factor or find its roots using the quadratic formula: \( r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} \). \( r = -1 \pm 3i \).Thus, the characteristic roots are \( r = i, -i, -1 + 3i, -1 - 3i \).

Step 4: Write the general solution to the homogeneous equation

From the roots found, the general solution to the differential equation \( y^{(4)} + 2y^{'''} + 11y^{''} + 2y' + 10y = 0 \) is:\[ y = c_1 \sin x + c_2 \cos x + e^{-x}(c_3 \cos(3x) + c_4 \sin(3x)) \],where \( c_1, c_2, c_3, \text{ and } c_4 \) are constants.

Step 5: Formulate the second-order homogeneous equation

You are given two solutions \( y_1 = 1 \) and \( y_2 = e^{-x} \). Assume a solution of the form \( y = e^{rx} \) and write the characteristic equation:\( (r - 0)(r + 1) = 0 \).The roots are \( r = 0 \) and \( r = -1 \), leading to the homogeneous part of the equation: \[ y^{''} + y' = 0 \].

Step 6: Incorporate the particular solution to find the nonhomogeneous equation

The nonhomogeneous equation has a particular solution \( y_p = \frac{1}{2}x^2 - x \). The complete equation is given by:\( y^{''} + y' = \frac{1}{2}x^2 - x \).Combine the homogeneous and particular parts to form the complete linear second-order differential equation.

Key Concepts

Characteristic Equation

In differential equations, the characteristic equation plays a crucial role in finding the solution. It emerges from assuming a solution of the form \( y = e^{rx} \) in a linear differential equation with constant coefficients. By substituting this assumed solution into the differential equation, you can transform the problem into a polynomial equation where the roots determine the structure of the general solution.
The characteristic equation is a polynomial, often called the auxiliary polynomial, whose degree is determined by the order of the differential equation being solved. For example, a fourth-order differential equation will lead to a fourth-degree polynomial.
Suppose we have a characteristic equation \( r^4 + 2r^3 + 11r^2 + 2r + 10 = 0 \). Solving this equation involves finding the polynomial's roots, which could be real or complex. These roots help in constructing the solution to the homogeneous differential equation.

Homogeneous Differential Equation

A homogeneous differential equation is one in which all terms are multiples of the function or its derivatives, set equal to zero. This type of equation is pivotal because finding its general solution is the first step in addressing more complex problems.
For instance, consider the equation \( y^{(4)} + 2y^{'''} + 11y^{''} + 2y' + 10y = 0 \). Since all terms involve the function \( y \) and its derivatives, the equation is homogeneous. By solving the characteristic equation derived from this, we find solutions related to the roots of the characteristic polynomial.
The general solution for this particular homogeneous differential equation would include sine and cosine terms corresponding to the complex roots \( r = i \) and \( r = -i \), and exponential combinations for other roots. The aim is to find the constants in the equation that satisfy initial or boundary conditions.

Nonhomogeneous Differential Equation

Nonhomogeneous differential equations include an additional term not related solely to the function or its derivatives. This term, also known as the nonhomogeneous part, is usually represented by a function of the independent variable.
Consider the differential equation \( y^{''} + y' = \frac{1}{2}x^2 - x \). Here, \( \frac{1}{2}x^2 - x \) is the nonhomogeneous term. The solution to a nonhomogeneous differential equation consists of two parts: the complementary solution, which solves the related homogeneous equation, and the particular solution, which addresses the nonhomogeneity.
Using known methods such as undetermined coefficients or variation of parameters, we find this particular solution. It combines with the complementary solution to form the complete solution of the nonhomogeneous equation.

Particular Solution

The particular solution of a nonhomogeneous differential equation specifically accounts for the nonhomogeneous part, or the external inputs of the system. Unlike the complementary solution, which addresses the equation's homogeneous portion, the particular solution is tied directly to the structure of the nonhomogeneity.
If given a nonhomogeneous differential equation like \( y^{''} + y' = \frac{1}{2}x^2 - x \), the particular solution could look like \( y_p = \frac{1}{2}x^2 - x \). This matches the nonhomogeneous term, crafted to satisfy the non-zero right-hand side of the equation.
To find the particular solution, techniques such as undetermined coefficients or variation of parameters are employed. These methods tailor the solution to the specific form of the nonhomogeneous term.

Roots of Polynomial Equations

The roots of polynomial equations are essential in solving differential equations because they dictate the form of the solution. These roots are values that satisfy the characteristic equation derived from a homogeneous differential equation.
For a given equation \( r^4 + 2r^3 + 11r^2 + 2r + 10 = 0 \), finding roots involves factoring the polynomial or using methods like the quadratic formula for simpler cases.
Roots can be real, leading to exponential growth or decay in solutions. Complex roots introduce sinusoidal behavior, i.e., sine and cosine functions. In our example, \( r = i, -i, -1 + 3i, -1 - 3i \) indicate a mix of oscillatory and exponential components in the general solution.