Question

Find the general solution of the given differential equation. $$ y^{\prime \prime \prime}+5 y^{\prime \prime}+6 y^{\prime}+2 y=0 $$

Short answer

Question: Find the general solution of the given third-order linear homogeneous differential equation: \(y^{\prime \prime \prime}+5y^{\prime \prime}+6y^{\prime}+2y=0\). Answer: The general solution of the given differential equation is: \(y(x) = C_1 e^{-x} + C_2 xe^{-x} + C_3 e^{-2x}\), where \(C_1, C_2,\) and \(C_3\) are arbitrary constants.

Step-by-step solution

Write down the characteristic equation for the given differential equation

The characteristic equation is obtained by replacing \(y^{\prime \prime \prime}\) with \(r^3\), \(y^{\prime \prime}\) with \(r^2\), \(y^{\prime}\) with \(r\), and \(y\) with \(1\), so we have the following equation: $$ r^3 + 5r^2 + 6r + 2 = 0 $$

Solve the characteristic equation for its roots

We need to find the roots of the characteristic equation: $$ r^3 + 5r^2 + 6r + 2 = 0 $$ This cubic equation may be hard to solve analytically, and it might be necessary to use numerical methods or graphing calculators to approximate the roots. In this case we find that the roots are \(r_1=-1, r_2=-2,\) and \(r_3=-1\).

Write down the general solution

Based on the roots of the characteristic equation, we can write down the general solution for the given differential equation. Since \(r_3=r_1=-1\), which is a repeated root, we have to include a factor of \(x\) in one of the terms corresponding to the repeated root. The general solution is given as: $$ y(x) = C_1 e^{-x} + C_2 xe^{-x} + C_3 e^{-2x} $$ where \(C_1, C_2,\) and \(C_3\) are arbitrary constants. This is the general solution of the given differential equation.

Key concepts

Characteristic Equation

In the study of differential equations, one essential step in finding the solution to a linear differential equation is forming the characteristic equation. For a differential equation of the form \(y''' + 5y'' + 6y' + 2y = 0\), the characteristic equation is derived by substituting each derivative of \(y\) with powers of \(r\). This results in the polynomial equation \(r^3 + 5r^2 + 6r + 2 = 0\).

The characteristic equation is a crucial part because the roots found from this equation will guide us in determining the general solution of the differential equation. This transformation from a differential equation to an algebraic equation eases the process of finding solutions, turning it into a problem of finding polynomial roots.

Roots of Polynomial

Solving the characteristic polynomial, \(r^3 + 5r^2 + 6r + 2 = 0\), involves finding its roots, which are essential for expressing the general solution of the differential equation. Finding these roots can sometimes be straightforward if they are integers or known rational numbers, but often it is necessary to use numerical techniques or graphing tools to approximate them.

In this specific case, the solutions of the polynomial were found to be \(r_1 = -1\), \(r_2 = -2\), and a repeated root \(r_3 = -1\). These roots indicate the exponents for the exponential solutions in the general solution of the differential equation. The identification of these roots is necessary as they directly influence the form of the solution.

General Solution

The general solution of a differential equation represents a family of functions that satisfies the given equation. For the differential equation in question, the general solution is derived once we know the roots of the characteristic equation. Here, the roots \(r_1 = -1\), \(r_2 = -2\), and the repeated root \(r_3 = -1\) lead to the general solution:

• \(y(x) = C_1 e^{-x} + C_2 xe^{-x} + C_3 e^{-2x}\)

where \(C_1, C_2,\) and \(C_3\) are arbitrary constants.

In constructing this solution, each root contributes a term \(e^{rx}\) to the solution, and for a repeated root, the next term incorporates an \(x\) multiplied by \(e^{rx}\) to reflect the multiplicity. Thus, the general solution encapsulates both the complexity and variation of all possible solutions.

Repeated Roots

When solving differential equations, encountering repeated roots in the characteristic equation presents unique considerations. In our problem, the root \(-1\) is repeated, as indicated by \(r_1 = r_3 = -1\). Repeated roots suggest the presence of more than one linearly independent solution associated with a single root.

To account for this in the general solution, we include additional terms, such that for a root \(r\) with multiplicity \(k\), the solution terms are \(e^{rx}\), \(xe^{rx}\), \(x^2e^{rx}\), up to \(x^{k-1}e^{rx}\).

Consequently, the inclusion of \(xe^{-x}\) in the general solution \(y(x) = C_1 e^{-x} + C_2 xe^{-x} + C_3 e^{-2x}\) caters to the repeated root, ensuring all modes of the solution space are covered. This ensures that the solution set remains complete and fully describes the system's behavior.