Question

Suppose the characteristic polynomial of \(a y^{\prime \prime}+b y^{\prime}+c y=0\) has complex conjugate roots \(\lambda \pm i \omega\). Use a method suggested by Exercise 22 to find a formula for the solution of $$ a y^{\prime \prime}+b y^{\prime}+c y=0, \quad y\left(x_{0}\right)=k_{0}, \quad y^{\prime}\left(x_{0}\right)=k_{1} $$

Short answer

Question: Given a linear second order differential equation with constant coefficients whose characteristic polynomial has complex conjugate roots and the initial conditions y(x0) = k0 and y'(x0) = k1, find a formula for the solution. Solution: The solution of the given problem is a function in the form: $$ y(x) = C e^{\lambda x} \left( \cos (\omega x) + D \sin (\omega x) \right) $$ where C and D are real constants determined by solving the system of equations: $$ k_0 = C e^{\lambda x_0} \left( \cos (\omega x_0) + D \sin (\omega x_0) \right) $$ $$ k_1 = \lambda C e^{\lambda x_0}\left( \cos (\omega x_0) + D \sin (\omega x_0) \right) + C e^{\lambda x_0} \omega \left(-\sin (\omega x_0) + D \cos (\omega x_0) \right) $$

Step-by-step solution

Write down the given information

We have the second order differential equation: $$ a y'' + b y' + c y = 0 $$ And its characteristic polynomial has complex conjugate roots \(\lambda \pm i \omega\). The given initial conditions are \(y(x_0) = k_0\) and \(y'(x_0) = k_1\).

Find the general solution

The general solution of our differential equation is a linear combination of its complex conjugate solutions: $$ y(x) = A e^{(\lambda + i \omega)x} + B e^{(\lambda - i \omega)x} $$ where A and B are complex constants.

Express the solution in terms of the real and imaginary parts

Using Euler's formula, we can rewrite the exponentials in terms of the real and imaginary parts: $$ y(x) = C e^{\lambda x} \left( \cos (\omega x) + D \sin (\omega x) \right) $$ where \(C = |A + B|\) and \(D = |\frac{A - B}{i}|\) are real constants.

Apply the initial conditions

To find C and D, we need to apply the initial conditions \(y(x_0) = k_0\) and \(y'(x_0) = k_1\). For the first initial condition \(y(x_0) = k_0\): $$ k_0 = C e^{\lambda x_0} \left( \cos (\omega x_0) + D \sin (\omega x_0) \right) $$ For the second initial condition \(y'(x_0) = k_1\), we first need to find the derivative \(y'(x)\): $$ y'(x) = \lambda C e^{\lambda x}\left( \cos (\omega x) + D \sin (\omega x) \right) + C e^{\lambda x} \omega \left(-\sin (\omega x) + D \cos (\omega x) \right) $$ Now substitute \(x = x_0\) and set it equal to \(k_1\): $$ k_1 = \lambda C e^{\lambda x_0}\left( \cos (\omega x_0) + D \sin (\omega x_0) \right) + C e^{\lambda x_0} \omega \left(-\sin (\omega x_0) + D \cos (\omega x_0) \right) $$

Solve for C and D

We now have a system of two equations to solve for C and D. This step may require dividing the equations or using other algebraic techniques depending on the values of \(\lambda, \omega, k_0, k_1, x_0\). Once C and D are found, the solution of the original problem is: $$ y(x) = C e^{\lambda x} \left( \cos (\omega x) + D \sin (\omega x) \right) $$

Key concepts

Characteristic Polynomial

When dealing with a second order differential equation, a powerful tool for finding the general solution is the characteristic polynomial. This polynomial arises from an algebraic counterpart to the differential equation by substituting the derivatives with powers of a variable (commonly 'r' or 'λ') and equating the expression to zero.

For a second order differential equation of the form \(ay'' + by' + cy = 0\), the characteristic equation is \(ar^2 + br + c = 0\). When the roots of this polynomial are complex, they occur as conjugate pairs \(λ ± iω\). These roots suggest the behavior of the solution: an exponential growth or decay (depending on the sign of λ) modulated by an oscillatory function (determined by ω). Understanding the characteristic polynomial helps us predict the form of the general solution, even before we calculate it explicitly.

Second Order Differential Equation

A second order differential equation involves the second derivative of a function and can often describe systems with inertia, such as mechanical vibrations or electrical circuits. The most general linear form is \(ay'' + by' + cy = 0\), where 'y' is the function of the variable 'x', and 'a', 'b', and 'c' are constant coefficients. The solutions to such equations are crucial in the modeling of physical systems.

To solve these equations, we look at the characteristic polynomial to determine the nature of the solutions. If the roots are complex conjugate, as in our exercise, the solution combines exponential and trigonometric functions to capture both the damping and oscillatory aspects of the system's behavior.

Euler's Formula

Euler's formula is a gem in the field of complex analysis that connects complex exponentials with trigonometric functions. It is represented as \(e^{ix} = \cos(x) + i\sin(x)\) for any real number 'x'. When used in the context of differential equations with complex roots, Euler's formula allows us to express complex exponentials as a combination of cosine and sine functions, which are real and easier to interpret in terms of physical phenomena.

The true value of this formula lies in its ability to simplify our solutions. In the given step-by-step solution, Euler's formula is applied to convert the general solution with complex exponentials into a form involving only real numbers, thus making it possible to apply real initial conditions effectively.

Initial Conditions

Solving a second order differential equation typically provides us with a general solution that includes two arbitrary constants. These constants are determined by using initial conditions, which specify the state of the system at a particular point, usually the start of our observation.

In our exercise, the initial conditions given are \(y(x_0) = k_0\) and \(y'(x_0) = k_1\). By substituting these into the general solution and its derivative, we can solve for the constants in the context of the problem. This step is crucial because it 'initializes' our general solution, tailoring it to a particular scenario and making the abstract, general solution into a concrete, particular one that describes a real-world situation.