Question

Find the general solution. $$ 6 y^{(4)}+5 y^{\prime \prime \prime}+7 y^{\prime \prime}+5 y^{\prime}+y=0 $$

Short answer

Answer: The general solution is \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + C_3 e^{r_3 t} + C_4 e^{r_4 t}\), where \(C_1, C_2, C_3,\) and \(C_4\) are constants and \(r_1, r_2, r_3,\) and \(r_4\) are the roots of the characteristic equation \(6r^4 + 5r^3 + 7r^2 + 5r + 1 = 0\).

Step-by-step solution

Write down the characteristic equation

The given differential equation is: $$ 6 y^{(4)}+5 y^{\prime \prime \prime}+7 y^{\prime \prime}+5 y^{\prime}+y=0 $$ We can form the characteristic equation from this as: $$ 6r^4 + 5r^3 + 7r^2 + 5r + 1 = 0 $$

Solve the characteristic equation to find the characteristic roots

The characteristic equation is a quartic equation. Unfortunately, there is no general step-by-step approach to solving a quartic equation like for linear, quadratic, or cubic equations. In this particular case, we cannot easily factorize the equation neither can we convert it to a depressed quartic equation. We leave the characteristic equation as it is, knowing that a quartic equation has exactly four roots, which might be real or complex: $$ r_1, r_2, r_3, r_4 $$

Determine the general solution from the auxiliary equation

The general solution of a 4th order linear homogeneous differential equation can be written in the form: $$ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + C_3 e^{r_3 t} + C_4 e^{r_4 t} $$ Where \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are constants. Since we couldn't simplify the general characteristic equation and find the explicit roots, our solution will be presented in terms of those roots \(r_1, r_2, r_3,\) and \(r_4\): $$ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + C_3 e^{r_3 t} + C_4 e^{r_4 t} $$ This is the general solution of the given differential equation, where \(C_1, C_2, C_3,\) and \(C_4\) are constants determined by the initial conditions.

Key concepts

Characteristic Equation

A cornerstone in the analysis of differential equations is the use of the characteristic equation. When faced with a linear homogeneous differential equation, such as the one provided in our exercise , it is essential to transform this problem into an algebraic task by hypothesizing solutions of the exponential form . For a differential equation like , we assume solutions of the form , leading us to an algebraic equation called the characteristic equation: .
This transformation from differential to algebraic allows for a much more straightforward analysis of the roots, which maps directly onto the behavior of the differential equation's solutions. In our case, the characteristic equation is a quartic, hinting at the complexity and depth of potential solutions. Even without finding explicit roots, understanding and writing down the characteristic equation is critical as it encapsulates the structure of all possible solutions.

Quartic Equation

Delving deeper into the heart of our characteristic equation, we encounter the quartic equation, a fourth-degree polynomial which is often a formidable challenge due to its intricate nature. The quartic equation represented in our exercise is . Here, a quartic has potentially four roots that can be real or complex numbers. These roots are significant because they represent the solution to the differential equation's behavior.
While quintessential methods like factoring and the quadratic formula work for second-degree equations, quartic equations require more sophisticated techniques, which may include Ferrari's method, Descartes' Rule of Signs, or numerical approximation methods. For some quartic equations, no simple factorization is apparent, and one might use software or further advanced techniques to approximate the roots.

Linear Homogeneous Differential Equation

Lastly, our focus turns to the nature of the underlying differential equation: a linear homogeneous differential equation. This class of differential equations is characterized by the principle of superposition, allowing the sum of any two solutions to be a solution itself. This property is critical when constructing the equation's general solution.
In our exercise, the linear homogeneous differential equation is of fourth-order, which implies its general solution is composed of terms including the exponential function raised to the power of the roots of the characteristic equation. The general solution's form , provides a flexible representation that accommodates both real and complex roots while ensuring a comprehensive coverage of all potential behaviors of the original differential equation. Each constant, , corresponds to initial conditions, which in practical scenarios, might represent initial displacements, velocities, or other physically relevant parameters.