Solving Second-Order Differential Equations
Different types for Solving Second-Order Differential Equations are:
- Analytical Method
- Numerical Method
Analytical Method
An analytical way of solving differential equations with such methods as undetermined coefficients and the method of variation of parameters is applied. The particular solution for an undetermined coefficients method is approximated in a particular form and the coefficients are derived by substituting the solution back into the differential equation.
Let’s consider an example to illustrate the undetermined coefficients method for solving second-order linear differential equations with constant coefficients:
Example: Solve the differential equation y′′- 3y′+2y = 2ex
Solution:
Step 1: Solve the associated homogeneous equation
y′′- 3y′ + 2y = 0
Characteristic equation is r2– 3r + 2=0, which has roots r1 = 1 and r2 = 2
Therefore, the general solution of the homogeneous equation is:
yc = c1ex + c2e2x
Step 2: Assume a particular solution of the form yp = Aex
Substitute yp into the original equation:
(Aex)′′ – 3(Aex)′ + 2(Aex) = 2ex
Simplify and solve for A:
Aex – 3Aex + 2Aex = 2ex
-Aex = 2ex
A = -2
Step 3: Write the general solution
y = yc + yp
y = c1ex + c2e2x – 2ex
Numerical Method
Numerical approaches are used for situations when the analytical solution is not possible. Approximate solutions are calculated by applying methods like Euler’s Technique and Runge’s Kutta Methods. The Euler Method applies a technique of partitioning the domain, to which the tangent lines then are used for the finite solution approximation purpose.
Runge-Kutta Methods are higher-order numerical techniques that form a particular type of approximation, known as the higher-order Runge-Kutta, by taking multiple intermediate steps.
Let’s consider an example to illustrate Euler’s method for solving second-order differential equations:
Example: Solve the initial value problem y′ = x + y, y(0) = 1 on the interval using Euler’s method with step size h = 0.25.
Solution:
Step 1: Discretize the interval using the step size h = 0.25:
x0 = 0, x1 = 0.25, x2 = 0.50, x3 = 0.75, x4 = 1.00
Step 2: Use Euler’s method to approximate the solution at each point:
yn+1 = yn + hf(xn, yn)
Where f(x, y) = x + y
y0 = 1
y1 = y0 + hf(x0, y0) = 1 + 0.25(0 + 1) = 1.25
y2 = y1 + hf(x1, y1) = 1 .25 + 0.25(0.25 + 1.25) = 1.625
y3 = y2 + hf(x2, y2) = 1.625 + 0.25(0.50 + 1.625) = 2.125
y4 = y3 + hf(x3, y3) = 2.125 + 0.25(0.75 + 2.125) = 2.75
Therefore, using Euler’s method with a step size of h = 0.25, the approximate solution at x = 1 is y ≈ 2.75.
Second Order Differential Equation
Differential equations of the second order, in mathematics are differential equations involving the second-order derivative of a function. Second Order Differential Equation involves the second-order derivative of a function, which is critical in providing accurate models of various real-world instances.
Table of Content
- What is Differential Equations?
- What are Second Order Differential Equations?
- Types of Second-Order Differential Equations
- Solutions of Second-Order Differential Equations
- Solving Second-Order Differential Equations
- Solving Homogeneous Second-Order Differential Equation
- Solving Non-Homogeneous Second-Order Differential Equations
- Second-Order Differential Equation Examples
- Applications of Second-Order Differential Equations