Second Order Differential Equation

A second-order differential equation is a differential equation in which the highest-order derivative is the second derivative of the dependent variable.

• Naturally represent physical systems involving acceleration, such as oscillations, vibrations, and motion.
• Require more involved solution methods and typically need initial conditions to arrive at a specific solution.

General form: a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = f(x)

Types

Homogeneous and Non-homogeneous Equations

The right side is equal to zero. Homogeneous equations have the form

a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = 0.

Non-homogeneous equations have an additional function on the right-hand side, such as f(x), making them:

 a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = f(x)

Linear Differential Equations

Linear differential equations are equations where the dependent variable and its derivatives appear only to the first power, having general form:

a(x)\frac{d^2 y}{dx^2} + b(x)\frac{dy}{dx} + c(x)y = f(x)

Second Order Differential Equation with Constant Coefficients

These have coefficients that are constants throughout the equation.

a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)

Second Order Differential Equation with Variable Coefficients

These are equations where the coefficient of differential equation is a variable.

a(x)\frac{d^2y}{dx^2} + b(x)\frac{dy}{dx} + c(x)y = f(x)

General Solution

The complete solution of a non-homogeneous second order differential equation consists of two parts complementary function (homogeneous solution) and the particular integral (non-homogeneous solution).

Complementary Function (CF): The complementary function is obtained by solving the associated homogeneous equation.

Example: For y′′−5y′+6y = 0, the complementary function is

y_c=C_1e^{2x}+C_2e^{3x}

Particular Integral (PI): The particular integral is a specific solution of the non-homogeneous equation.

Example: For y′′+y = x a particular integral is determined separately.

Complete Solution: y = CF+PI or y = y_c+y_p

Solving Homogeneous Second-Order Differential Equation

For a homogeneous second-order differential equation of the form a d²y/dx² + b dy/dx + cy =0, where a,b, and c are constants:

Find the characteristic equation by substituting y = e^rx into the equation and simplifying.

• Solve the characteristic equation to find the roots r₁ and r₂​.
• If the roots are real and distinct, the general solution is y = c₁e^r1x + c₂e^r2x.
• If the roots are complex conjugates, the general solution is y = e^αx{c₁cos⁡(βx) + c₂sin⁡(βx)], where α and β are determined from the roots.
• If the roots are repeated, the general solution is y = (c₁+c₂x)e^rx, where r is the repeated root.

Solving Non-Homogeneous Second-Order Differential Equations

For non-homogeneous second-order differential equation of the form a d²y/dx² + b dy/dx + cy = f(x), where a,b, and c are constants, and f(x) is a function of x, we can use the method of undetermined coefficients or the method of variation of parameters.

• Solve the associated homogeneous equation to find the general solution y_c​.
• Find a particular solution y_p​ using the method of undetermined coefficients or variation of parameters.
• General solution of the non-homogeneous equation is y = y_c + y_p.

Methods for Solving

Method 1: Characteristic Equation Method

Used for linear differential equations with constant coefficients.

Steps to solve

1. Write the differential equation.
2. Form the characteristic equation.
3. Solve for the roots.
4. Construct the complementary function.

Example: y′′ − 5y′ + 6y = 0

Characteristic equation: m²−5m+6 = 0

(m-2)(m-3) = 0

Roots: m = 2, 3

Solution: y=C_1e^{2x}+C_2e^{3x}

Method 2: Undetermined Coefficients Method

Used when the forcing function is polynomial, exponential, sine, or cosine.

Example: y′′−y = e^x

Assume, y_p = Axe^x

Substitute into the equation and determine A.

Method 4: Cauchy-Euler Method

Applicable to equations of the form: ax²y′′+bxy′+cy = 0

Substitution: y = x^m

Example: x²y′′−3xy′+4y = 0

Substituting y = x^m gives an algebraic equation in m.

Applications

• Vibrations and Oscillations: Used to describe the motion of springs, pendulums, and other vibrating systems.
• Electrical Circuits: Applied in analyzing RLC circuits and signal transmission in electrical engineering.
• Population Dynamics: Helps model population growth and decline considering multiple influencing factors.
• Control Systems: Used in designing feedback and automation systems to ensure stability and desired performance.
• Structural Engineering: Helps analyze the behavior of structures such as bridges and buildings under various loads and forces.

Solved Examples

Example 1: Solve: \frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0

Step 1: Form the characteristic equation: m²−5m+6 = 0

Step 2: Factor the equation (m−2)(m−3)=0

Thus, the roots are: m=2,  3

Step 3: Write the general solution

Since the roots are distinct and real, the solution is: y=C_1e^{2x}+C_2e^{3x}

Example 2: Solve: \frac{d^2y}{dx^2}-4\frac{dy}{dx}+4y=0

Step 1: Form the characteristic equation : m²−4m+4 = 0

Step 2: Factor the equation: (m−2)2 = 0

Thus, the repeated root is: m = 2

Step 3: Write the general solution

For repeated roots, the solution is: y=(C_1+C_2x)e^{mx}

Substituting m = 2,

y=(C_1+C_2x)e^{2x}

Practice Problems

1. Solve: \frac{d^2y}{dx^2}-7\frac{dy}{dx}+12y=0

2. Solve: \frac{d^2y}{dx^2}+9y=0

3. Solve: \frac{d^2y}{dx^2}-6\frac{dy}{dx}+9y=0

4. Solve: \frac{d^2y}{dx^2}+4\frac{dy}{dx}+4y=0

5. Solve: \frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=0