Question

Find the general solution of the given differential equation. $$ y^{\mathrm{iv}}-5 y^{\prime \prime}+4 y=0 $$

Short answer

Question: Find the general solution of the given fourth-order linear homogenous differential equation with constant coefficients: \(y^{''''} - 5y^{''} + 4y = 0\). Answer: The general solution of the given differential equation is: $$ y(x) = c_1 e^x + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x}. $$ where \(c_1, c_2, c_3,\) and \(c_4\) are arbitrary constants determined by the initial conditions.

Step-by-step solution

Write down the characteristic equation from the given differential equation.

To do this, we replace the derivatives with their corresponding powers of the variable "r". The characteristic equation for the differential equation $$ y^{''''} - 5y^{''} + 4y = 0 $$ becomes $$ r^4 - 5r^2 + 4 = 0. $$

Solve for the roots of the characteristic equation.

The characteristic equation is a quadratic equation in terms of \(r^2\): $$ (r^2)^2 - 5(r^2) + 4 = 0. $$ Now, let \(a = r^2\). We have the quadratic equation $$ a^2 - 5a + 4 = 0. $$ The roots of this quadratic equation can be found by factoring: $$ (a-1)(a-4) = 0. $$ This gives us the roots \(a = 1\) and \(a = 4\). To find the corresponding roots of the original characteristic equation, we substitute back the variable "r", representing the original roots: \(r^2 = 1\) and \(r^2 = 4\). For \(r^2 = 1\), we have \(r = \pm 1\). For \(r^2 = 4\), we have \(r = \pm 2\). The roots of the characteristic equation are \(r_1 = 1, r_2 = -1, r_3 = 2,\) and \(r_4 = -2\).

Write down the general solution using the roots from Step 2.

Since we have found four distinct roots for the characteristic equation, the general solution of the given differential equation can be written as the linear combination of the exponential functions corresponding to the roots: $$ y(x) = c_1 e^{1x} + c_2 e^{-1x} + c_3 e^{2x} + c_4 e^{-2x}. $$ where \(c_1, c_2, c_3,\) and \(c_4\) are arbitrary constants determined by the initial conditions. So, the general solution of the given differential equation is: $$ y(x) = c_1 e^x + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x}. $$

Key concepts

Characteristic Equation

In the world of differential equations, a characteristic equation is a special polynomial derived from a linear differential equation with constant coefficients. This equation is crucial because it helps us find the particular solutions that will satisfy the differential equation. To form the characteristic equation, we replace each derivative term in the differential equation with powers of a variable, often denoted as "\(r\)". For instance, in the given differential equation \(y^{\text{iv}}-5y^{\prime\prime}+4y=0\), the characteristic equation emerges as \(r^4 - 5r^2 + 4 = 0\). The idea is converting the problem from a differential one to an algebraic one, making it easier to handle. By solving this characteristic polynomial, we find the roots that guide us to the solution of the original differential equation.

General Solution

The general solution of a differential equation encompasses all possible solutions to the equation, highlighted as a result of the diverse roots from the characteristic equation. When you have a differential equation with constant coefficients and distinct roots, the general solution can be expressed as a linear combination of exponential functions. Each term in this linear combination corresponds to one root of the characteristic equation. In this case, for the roots \(r_1 = 1, r_2 = -1, r_3 = 2,\) and \(r_4 = -2\), the general solution is built using expressions of the form \(e^{rx}\). Therefore, the solution is:

• \( y(x) = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x} \)

Here, \(c_1, c_2, c_3,\) and \(c_4\) are arbitrary constants that would be resolved using initial conditions. These express all possible solutions, enabling flexibility to match specific outset scenarios.

Exponential Function

Exponential functions form the backbone of solutions to many differential equations, especially those with constant coefficients. An exponential function is expressed in the form \(e^{rx}\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718, and \(r\) and \(x\) are variables, with \(r\) often being a root from the characteristic equation. These functions possess unique properties:

• They grow initially fast when \(r > 0\), and decay similarly when \(r < 0\).
• Their derivatives are proportional to themselves, making them highly suitable for solutions to differential equations.

When solving our differential equation \(y^{\text{iv}} - 5y^{\prime\prime} + 4y = 0\), each exponential term in the general solution corresponds to one of the distinct roots found. These exponential functions bring structure and completeness to the solution, allowing it to cater to various initial conditions and behaviors.