Question

In each exercise, you are given the general solution of $$ y^{(4)}+a_{3} y^{\prime \prime \prime}+a_{2} y^{\prime \prime}+a_{1} y^{\prime}+a_{0} y=0, $$ where \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\) are real constants. Use the general solution to determine the constants \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\). [Hint: Construct the characteristic equation from the given general solution.] $$ y(t)=c_{1} e^{-t} \sin t+c_{2} e^{-t} \cos t+c_{3} e^{t} \sin t+c_{4} e^{t} \cos t $$

Short answer

Given the general solution of the fourth-order homogeneous linear differential equation: $$ y(t)=c_{1} e^{-t} \sin t+c_{2} e^{-t} \cos t+c_{3} e^{t} \sin t+c_{4} e^{t} \cos t $$ We found the constants \(a_{3}, a_{2}, a_{1}\), and \(a_{0}\) to be \(0, 2, 0,\) and \(2\) respectively. Therefore, the corresponding differential equation is: $$ y^{(4)} + 2y^{\prime \prime} + 2y = 0 $$

Step-by-step solution

Identify the given general solution

We are given the general solution of the DE as: $$ y(t)=c_{1} e^{-t} \sin t+c_{2} e^{-t} \cos t+c_{3} e^{t} \sin t+c_{4} e^{t} \cos t $$

Compute the derivatives of the general solution

We need to calculate the first, second, third, and fourth derivatives to substitute into the DE: First derivative: $$ y'(t) = -c_{1} e^{-t} \sin t - c_{2} e^{-t} \cos t + c_{1} e^{-t} \cos t - c_{2} e^{-t} \sin t + c_{3} e^{t} \sin t + c_{4} e^{t} \cos t + c_{3} e^{t} \cos t - c_{4} e^{t} \sin t $$ Second derivative: $$ y''(t) = 2c_{1} e^{-t} \sin t + 2c_{2} e^{-t} \cos t - 2c_{1} e^{-t} \cos t + 2c_{2} e^{-t} \sin t + 2c_{3} e^{t} \sin t - 2c_{4} e^{t} \cos t - 2c_{3} e^{t} \cos t - 2c_{4} e^{t} \sin t $$ Third derivative: $$ y'''(t) = -3c_{1} e^{-t} \sin t - 3c_{2} e^{-t} \cos t + 3c_{1} e^{-t} \cos t - 3c_{2} e^{-t} \sin t - 3c_{3} e^{t} \sin t + 3c_{4} e^{t} \cos t + 3c_{3} e^{t} \cos t + 3c_{4} e^{t} \sin t $$ Fourth derivative: $$ y^{(4)}(t) = 4c_{1} e^{-t} \sin t + 4c_{2} e^{-t} \cos t - 4c_{1} e^{-t} \cos t + 4c_{2} e^{-t} \sin t + 4c_{3} e^{t} \sin t - 4c_{4} e^{t} \cos t - 4c_{3} e^{t} \cos t - 4c_{4} e^{t} \sin t $$

Substitute the derivatives into the DE

Plug the derivatives into the given DE: $$ 4c_{1} e^{-t} \sin t + 4c_{2} e^{-t} \cos t - 4c_{1} e^{-t} \cos t + 4c_{2} e^{-t} \sin t + 4c_{3} e^{t} \sin t - 4c_{4} e^{t} \cos t - 4c_{3} e^{t} \cos t - 4c_{4} e^{t} \sin t + a_{3}(-3c_{1} e^{-t} \sin t - 3c_{2} e^{-t} \cos t + 3c_{1} e^{-t} \cos t - 3c_{2} e^{-t} \sin t - 3c_{3} e^{t} \sin t + 3c_{4} e^{t} \cos t + 3c_{3} e^{t} \cos t + 3c_{4} e^{t} \sin t) + a_{2}(2c_{1} e^{-t} \sin t + 2c_{2} e^{-t} \cos t - 2c_{1} e^{-t} \cos t + 2c_{2} e^{-t} \sin t + 2c_{3} e^{t} \sin t - 2c_{4} e^{t} \cos t - 2c_{3} e^{t} \cos t - 2c_{4} e^{t} \sin t) + a_{1}(-c_{1} e^{-t} \sin t - c_{2} e^{-t} \cos t + c_{1} e^{-t} \cos t - c_{2} e^{-t} \sin t + c_{3} e^{t} \sin t + c_{4} e^{t} \cos t + c_{3} e^{t} \cos t - c_{4} e^{t} \sin t)+ a_{0}(c_{1} e^{-t} \sin t+c_{2} e^{-t} \cos t+c_{3} e^{t} \sin t+c_{4} e^{t} \cos t) = 0 $$

Construct the characteristic equation

From the DE, we can construct the characteristic equation: $$ r^4 + a_{3}r^3 + a_{2}r^2 + a_{1}r + a_{0} = 0 $$ The roots of the characteristic equation are the exponents of the functions in the general solution. In our case, the roots are \(r_1 = -1 + i, r_2 = -1 - i, r_3 = 1 + i,\) and \(r_4 = 1 - i\).

Solve for the constants

Plug the roots into the characteristic equation and form a system of equations: 1. \((1+(-1+i))^4 + a_3(1+(-1+i))^3 + a_2(1+(-1+i))^2 + a_1(1+(-1+i)) + a_0 = 0\) 2. \((1+(-1-i))^4 + a_3(1+(-1-i))^3 + a_2(1+(-1-i))^2 + a_1(1+(-1-i)) + a_0 = 0\) 3. \((1+(1+i))^4 + a_3(1+(1+i))^3 + a_2(1+(1+i))^2 + a_1(1+(1+i)) + a_0 = 0\) 4. \((1+(1-i))^4 + a_3(1+(1-i))^3 + a_2(1+(1-i))^2 + a_1(1+(1-i)) + a_0 = 0\) Solving this system of equations, we find the unknown coefficients: $$ a_3 = 0, \quad a_2 = 2, \quad a_1 = 0, \quad a_0 = 2 $$ Therefore, the given general solution corresponds to the differential equation: $$ y^{(4)} + 2y^{\prime \prime} + 2y = 0 $$

Key concepts

Characteristic Equation

In the context of higher-order differential equations, constructing a characteristic equation is an essential step in solving homogeneous linear differential equations. The characteristic equation is created by replacing each derivative in a differential equation with powers of a variable, typically denoted by \( r \). This forms a polynomial equation whose roots are critical to finding the solution of the differential equation. For a fourth-order differential equation, this polynomial will be of degree four. For example, a typical fourth-order differential equation:\[ y^{(4)} + a_{3}y^{'''} + a_{2}y^{''} + a_{1}y' + a_{0}y = 0 \]leads to the characteristic equation:\[ r^4 + a_{3}r^3 + a_{2}r^2 + a_{1}r + a_{0} = 0 \] The characteristic equation allows us to determine the structure of the general solution by analyzing its roots. Each root corresponds to a term in the solution. Properly understanding how to form and solve the characteristic equation is a key skill in mastering differential equations.

Complex Roots

When solving the characteristic equation, encountering complex roots is a common scenario. Complex roots arise when the discriminant of the polynomial is negative, leading to non-real numbers. In the context of differential equations, if a characteristic equation has complex roots, they will usually appear in conjugate pairs such as \(r_1 = a + bi\) and \(r_2 = a - bi\). The presence of complex roots affects the form of the solution. Instead of simple exponentials, the solution will incorporate combinations of sines and cosines. The general form of a solution involving complex roots is:\[ y(t) = e^{at}(C_1 \, \cos(bt) + C_2 \, \sin(bt)) \]This formulation leverages Euler’s formula, which bridges the gap between complex exponentials and trigonometric functions. Recognizing and handling complex roots is crucial because they introduce oscillatory components into the solution, representing phenomena like waves and oscillations in physical systems.

Homogeneous Linear Differential Equations

Homogeneous linear differential equations have a characteristic that the output (the dependent variable, typically \(y\)) is proportional to its inputs (the independent variable and its derivatives). The defining feature of these equations is that setting the entire expression to zero yields the structure:\[ a_ny^{(n)} + a_{n-1}y^{(n-1)} + \, ... \, + a_1y' + a_0y = 0 \]The solution of a homogeneous linear differential equation hinges entirely on the characteristic equation. Its homogeneity means there are no external forces or inputs acting on the system; instead, it's the internal properties of the system's equation that drive its behavior. It is important to note that if a particular solution exists, any constant multiple or linear combination of solutions is also valid due to the linear nature of the equation.

General Solution of Differential Equations

The general solution of a differential equation involves finding all possible solutions that satisfy the given differential equation. For higher-order equations, especially those with constant coefficients, the general solution fundamentally depends on the roots of the characteristic equation. When all roots are distinct and real, the solution is purely made of exponential functions. If roots are complex, as discussed before, the solution incorporates trigonometric functions. Should any roots be repeated, the solution must adapt by multiplying terms by \( t \) as needed to maintain linear independence. In essence, the general solution is a comprehensive expression consisting of:

• Real exponentials for real roots
• Trigonometric functions for complex roots
• Polynomial terms in cases of repeated roots

For a given problem, we create a superposition of all individual solutions from each distinctive root of the characteristic equation. This combination constitutes the most inclusive form of the solution, encapsulating the behavior of the differential equation under any initial conditions.