Question

Find a fundamental set of solutions. $$ \left(4 D^{2}+1\right)^{2}\left(9 D^{2}+4\right)^{3} y=0 $$

Short answer

Answer: The fundamental set of solutions for the given differential equation is: $$ \left\{ \cos \frac{x}{2} + i \sin \frac{x}{2}, \cos \frac{x}{2} - i \sin \frac{x}{2}, \cos \frac{2x}{3} + i \sin \frac{2x}{3}, \cos \frac{2x}{3} - i \sin \frac{2x}{3} \right\} $$

Step-by-step solution

Factor out the Characteristic Polynomial

The provided differential equation is \((4D^2 + 1)^2(9D^2 + 4)^3y = 0\). We want to rewrite it in the concept of characteristic polynomial. Use the fact that, given a linear homogeneous differential equation $$ a_ny^{(n)}(x) + a_{n-1}y^{(n-1)}(x)+\cdots+a_1y'(x)+a_0y(x)=0, $$ the characteristic polynomial is $$ p(D)=a_nD^{n}+a_{n-1}D^{n-1}+\cdots+a_1D+a_0. $$ Equating the given equation to the form of a linear homogeneous differential equation, we obtain: $$ [ (4D^2 + 1)^2(9D^2 + 4)^3]=0 $$

Turn Each Parenthesis into a Quadratic Equation

Expand the equation from the previous step and focus on each parenthesis separately. 1) \((4D^2 + 1)^2 = 0 \Rightarrow 4D^2 + 1 = 0\). To solve for D, do the following: $$ D^2 = - \frac{1}{4} \Rightarrow D_1, D_2 = \pm \frac{i}{2}, $$ where \(i=\sqrt{-1}\). 2) \((9D^2 + 4)^3 = 0 \Rightarrow 9D^2 + 4 = 0\). To solve for D, do the following: $$ D^2 = - \frac{4}{9} \Rightarrow D_3, D_4 = \pm \frac{2i}{3}. $$

Build the Fundamental Solution Set

Now that we have the roots \(D_1, D_2, D_3\), and \(D_4\), we can build the fundamental solution set for the given differential equation. Since the roots are complex, we will have complex solutions that come in pairs. The general solution in this case is given by: $$ y(x) = C_1 e^{D_1 x} + C_2 e^{D_2 x} + C_3 e^{D_3 x} + C_4 e^{D_4 x} $$ Substituting the roots: $$ y(x) = C_1 e^{\frac{i}{2} x} + C_2 e^{-\frac{i}{2} x} + C_3 e^{\frac{2i}{3} x} + C_4 e^{-\frac{2i}{3} x} $$ Finally, applying Euler's formula: $$ y(x) = C_1 \left(\cos \frac{x}{2} + i \sin \frac{x}{2}\right) + C_2 \left(\cos \frac{x}{2} - i \sin \frac{x}{2}\right) + C_3 \left(\cos \frac{2x}{3} + i \sin \frac{2x}{3}\right) + C_4 \left(\cos \frac{2x}{3} - i \sin \frac{2x}{3}\right) $$ Thus, the fundamental set of solutions is: $$ \left\{ \cos \frac{x}{2} + i \sin \frac{x}{2}, \cos \frac{x}{2} - i \sin \frac{x}{2}, \cos \frac{2x}{3} + i \sin \frac{2x}{3}, \cos \frac{2x}{3} - i \sin \frac{2x}{3} \right\} $$

Key concepts

Characteristic Polynomial

Understanding the characteristic polynomial is key to solving higher-order linear homogeneous differential equations. When we're given an equation such as \( (4D^2 + 1)^2(9D^2 + 4)^3y = 0 \), we're looking at a differential operator acting on the function \(y(x)\). To find the roots of the equation, that is, the values of \(D\) that make the equation true, we turn to the characteristic polynomial. This polynomial is derived from the differential equation by replacing the differentiation operator \(\frac{d}{dx}\) with a variable, usually denoted \(D\).

For our equation, the characteristic polynomial is the expanded form of \( (4D^2 + 1)^2(9D^2 + 4)^3 \), which we equate to zero. Solving this polynomial will give us the roots or eigenvalues that are crucial for building the solution to the differential equation. These roots inform us about the behavior of the solutions, such as whether they oscillate (when the roots are imaginary) or whether they grow or decay exponentially (when the roots are real).

Linear Homogeneous Differential Equation

A linear homogeneous differential equation is one where the highest degree of any term is the first degree, and there are no products of the function \(y\) and its derivatives. The equation is homogeneous because there is no standalone function of \(x\); all terms involve the function \(y\) we're trying to solve for. Thus, all solutions to a homogeneous equation are multiples of each other, forming a vector space.

In the context of our problem, the structure of the given equation, once the characteristic polynomial is factored out, is indeed linear and homogeneous. When dealing with linear homogeneous equations, one crucial step is to determine the fundamental set of solutions, which are the building blocks for the general solution. These are obtained by solving the characteristic polynomial for its roots, as illustrated in the exercise.

Euler's Formula

Euler's formula, an essential tool in differential equations, connects complex exponentials to trigonometric functions. The formula states that for any real number \(x\), \( e^{ix} = \cos(x) + i \sin(x) \). This relationship is especially useful when solving differential equations with complex roots.

When we establish the roots of the characteristic polynomial as complex numbers, we use Euler's formula to express the complex exponentials as combinations of sine and cosine functions. This helps to create real-valued functions as solutions to the differential equation, which are more meaningful in most physical applications. For instance, in the given exercise, Euler's formula allows us to re-express the solutions involving \(\frac{i}{2}x\) and \(\frac{2i}{3}x\) using trigonometric functions, making the complex nature of the roots palatable and converting them into a form that can be interpreted in a real-world context.