Second Order Linear Differential Equation

The second-order linear differential equation involves the second derivative of a function. It is crucial in many areas of physics and engineering, such as mechanical vibrations, electrical circuits, and heat conduction.

The general form of a second-order linear differential equation is:

a(x)y" + b(x)y" + c(x)y = f(x)

• Where and b are functions of the independent variable x.

Types of Second-Order Linear Differential Equations

There are two types of second-order linear differential equations: Homogeneous equations and Non-Homogeneous equations.

Homogeneous Equation

A homogeneous equation in the context of differential equations has the general form:

A(x)d²y/dx² + B(x)dy/dx + C(x)y = 0

• Where A(x), B(x), and C(x) are functions of the independent variable x.

For simplicity, let's assume that A(x) = A, B(x) = B, and C(x) = C are constant. The equation becomes:

Ay" + By' + Cy = 0

Key Properties of Homogeneous Equations

• Superposition Principle: If g(x) is a solution of the homogeneous equation, then any constant multiple of g(x).
• Sum of Solutions: If h(x) and g(x) are both solutions, then h(x) + g(x) is also a solution.

General Solution of homogeneous equation:

To solve the equation Ay" + By' + C = 0 we follow thses steps:

• Assume a solution of the form y = e^rx , where r is constant.
• Substitute into the differential equation: Ar²e^rx + Bre^rx+Ce^rx=0
• Factor out e^rx ( which is never zero) : Ar² + Br + C = 0

Thus, the characterstic equation is: Ar² + Br + C = 0. We now solve this quadratic equation to find the roots r₁​ and r₂​.

Case 1: Two Real and Distinct Roots

If the roots of the characteristic equation are real and distinct, say r₁ and r₂, then the general solution is:

y = c₁e^rx + c₂e^rx

Case 2: Complex conjugate Roots

If the rooots are complex , say r₁ = a + b_i, where i is imaginary unit, the general solution take the form:

y = e^ax(c₁​cos(bx) + c_2​sin(bx))

This result uses Euler's formula e^ix= cos⁡(x) + isin⁡(x).

Case 3: Repeated Real Roots

If the characteristic equation has repeated real roots, say r₁ = r₂=r, then the general solution is:

y = (c₁ + c₂x)e^rxy

This is because the solution to the repeated root case involves multiplying by x to ensure linear independence.

Non-Homogeneous Second-Order Linear Differential Equations

A non-homogeneous second-order linear differential equation has the form:

A(x)d²y/dx² + B(x)dy/dx + C(x)y = g(x)

Where g(x) is a non-zero function, and the goal is to find a solution to this equation. The solution consists of two parts:

1. Complementary Function (C.F.): The solution to the associated homogeneous equation Ay′′+ By′+ Cy = 0.
2. Particular Integral (P.I.): A particular solution to the non-homogeneous equation.

The general solution is:

y(x) = C.F. + P.I

Finding the Particular Integral (P.I.):

The method of undetermined coefficients is often used to find the particular integral. The form of the P.I. depends on the function g(x). For example:

• If g(x) is a polynomial, exponential, or sinusoidal function, the form of the P.I. is chosen based on the form of g(x).
• If g(x) involves terms similar to the solutions of the homogeneous equation, we multiply by an additional factor of x to ensure linear independence.

Example 1: Solve y" + 5y' + 6y = 0

Solution:

Assume y = e^rx

Putting this in the equation, we finally get:

r² + 5r + 6 = 0
(r+2)(r+3) = 0
r = (-2) OR r = (-3)

So, r₁ = (-2) and r₂ = (-3)

Since both are real the general solution will be:

y = c₁e^(-2x) + c₂e^(-3x)

Solved Questions on Second Order Linear Differential Equation

Question 1: Solve y" + y' + y = 0

Solution:

Assume y = r^rx

and solve for 'r'. Your 'r' will look something like this:

r₁ = (-1/2) + i(-\sqrt(3)/2
and r₂ = (-1/2) - i-\sqrt(3)/2
So,
a₁ = (-1/2)
and a₂ = (\sqrt(3)/2)

Hence the general solution will look like this:

y = e^(-x/2)(c₁cos(x√(3)/2) + c₂sin(x√(3)/2))

Question 2: Solve y" + 4y' + 4y = 0

Solution:

Assume y = r^rx and solve for 'r'. The 'particular solution will be:

y = ce^2x

Assume

y = p(x)e^2x

Putting it in the differential equation will give you:

p" = 0 which implies
p'= c₂ which again implies
p = c₁x + c₂

Hence the general solution will be:

y = (c₁x + c₂)e^2x

Question 3: Solve:   y′′− 4y′+ 4y = 0.

Solution:

Assume y = r^rx and solve for 'r'. The 'particular solution will be:

r² − 4r + 4 = 0

(r - 2)² = 0

Repeated Root r = 2

General solution:

y = ( C₁ + C₂x)e^2x

Question 4: Solve:   y′′+ 2y′+ 5y = 0.

Solution:

Assume y = r^rx and solve for 'r'. The 'particular solution will be:

Characteristic:

r² + 2r + 5 = 0

Discriminant Δ = 4 − 20 = − 16

r =- 2 ± 4i/2 = -1 ± 2i .

So a = - 1, b = 2

General real solution:

y = e^-x (C₁ cos 2x + C₂ sin 2x)

Unsolved Questions on Second Order Linear Differential Equation

Question 1 : Solve the differential equation: y′′ − 6y′ + 9y = 0.

Question 2: Solve the non-homogeneous differential equation: y′′ + 2y′ + y = x².

Question 3: Solve the differential equation: y′′ + 9y = sin⁡(3x)

Question 4: Solve the differential equation: y′′ + 4y′ + 13y = 0.