Question

Find the general solution of the given differential equation. $$ y^{v}-3 y^{\mathrm{iv}}+3 y^{\prime \prime \prime}-3 y^{\prime \prime}+2 y^{\prime}=0 $$

Short answer

Answer: The general solution for the given differential equation is: $y(x) = C_1 + C_2 e^x + C_3 x e^x + C_4 x^2 e^x + C_5 e^{2x}$.

Step-by-step solution

Write down the given differential equation

The given differential equation is $$ y^{(5)}-3 y^{(4)}+3 y^{\prime \prime \prime}-3 y^{\prime \prime}+2 y^{\prime}=0 $$ where \(y^{(5)}\) represents the 5th derivative of \(y\) with respect to \(x\), \(y^{(4)}\) represents the 4th derivative, and so on.

Formulate the characteristic equation

For a linear homogeneous differential equation with constant coefficients, the characteristic equation is derived by replacing every derivative with a power of \(r\). Therefore the characteristic equation will be: $$ r^5 - 3r^4 + 3r^3 - 3r^2 + 2r = 0 $$

Solve the characteristic equation

First, notice that we can factor out \(r\) from each term in the equation. This gives us: $$ r(r^4 - 3r^3 + 3r^2 -3r + 2) = 0 $$ So we have found one root: \(r = 0\). Now we need to find the roots of the remaining quartic equation: $$ r^4 - 3r^3 + 3r^2 -3r + 2 = 0 $$ By trial and error or using a quartic equation solver, we find the roots to be: \(r = 1, 1, 1,\) and \(2\).

Write down the general solution

The general solution of a linear homogeneous differential equation with constant coefficients is a linear combination of exponentials where the exponents are the roots of the characteristic equation. Since the root \(r=1\) has multiplicity 3, we must include a polynomial factor of \(1, x, x^2\) for each exponent term: $$ y(x) = C_1 e^{0x} + C_2 e^{1x} + C_3 xe^{1x} + C_4 x^2 e^{1x} + C_5 e^{2x} $$

Simplify the general solution

Our general solution can be simplified slightly since \(e^{0x} = 1\). Therefore, the final general solution is: $$ y(x) = C_1 + C_2 e^x + C_3 x e^x + C_4 x^2 e^x + C_5 e^{2x} $$ This is the general solution for the given differential equation.

Key concepts

Characteristic Equation

To solve a higher-order linear homogeneous differential equation with constant coefficients, the first step is to find what is known as the characteristic equation. This is essentially a polynomial equation used to find the solution of differential equations. It is derived by substituting each derivative with a power of a variable, typically denoted as \( r \). For example, in the given exercise, the characteristic equation is formed by replacing each derivative in the order of the differential equation with its respective power of \( r \):

• 5th derivative becomes \( r^5 \)
• 4th derivative becomes \( r^4 \)
• 3rd derivative becomes \( r^3 \)
• 2nd derivative becomes \( r^2 \)
• 1st derivative becomes \( r \)

Plugging these into the equation, we obtain the polynomial \( r^5 - 3r^4 + 3r^3 - 3r^2 + 2r = 0 \). Solving this characteristic polynomial is key to finding the solution to the differential equation.

Roots of Polynomial

Solving the characteristic polynomial involves finding the roots of the equation. A root of a polynomial is a value that satisfies the equation, making it equal to zero. These roots are crucial because they help us determine the form of the general solution. In the exercise, by factoring the characteristic equation \( r(r^4 - 3r^3 + 3r^2 - 3r + 2) = 0 \), we immediately find one root: \( r = 0 \). This is because when \( r = 0 \), the entire expression becomes zero.Next, we need to solve the quartic part \( r^4 - 3r^3 + 3r^2 - 3r + 2 = 0 \). Through methods like trial and error or using a polynomial solver, we find that the remaining roots are \( r = 1, 1, 1, \) and \( 2 \). Note that \( r = 1 \) is a repeated root, which is essential for determining the general solution.

Linear Homogeneous Equations

A linear homogeneous differential equation is characterized by the sum of derivative terms set to zero, like the equation we solved: \( y^{(5)} - 3y^{(4)} + 3y^{(3)} - 3y^{(2)} + 2y^{(1)} = 0 \). "Linear" denotes that each term is a linear function of the derivatives, and "homogeneous" means that every term equals zero, which allows solutions to be scaled and combined.The significance of these equations lies in their simplicity and structure, enabling the formation of characteristic equations. The structure, determined by consistently applied rules, simplifies finding solutions. The solutions to a homogeneous differential equation are not unique; however, they can be expressed as linear combinations of functions derived from the characteristic roots.

General Solution

The general solution for a linear homogeneous differential equation with constant coefficients involves constructing a combination of exponential functions whose exponents comprise the roots of the characteristic equation. The roots we found are \( r = 0, 1, 1, 1, \) and \( 2 \). Each root translates into the following components in the general solution:

• For \( r = 0 \), the factor is \( C_1 e^{0x} = C_1 \).
• For \( r = 1 \) (triple root), factors are \( C_2 e^x, C_3 x e^x, C_4 x^2 e^x \) reflecting the multiplicity.
• For \( r = 2 \), the factor is \( C_5 e^{2x} \).

Thus, the complete general solution can be expressed as \( y(x) = C_1 + C_2 e^x + C_3 x e^x + C_4 x^2 e^x + C_5 e^{2x} \). This solution represents all possible solutions to the differential equation, with \( C_1, C_2, C_3, C_4, \) and \( C_5 \) being arbitrary constants that depend on initial conditions.