How to Solve a Second Order Differential Equation with Formula and Examples
A differential equation is an equation of a function and one or more derivatives which may be of first degree or more.
Differential Equations are of the form: d2y/dx2 + p dy/dx + qy = 0.
Differential Equations might be of different orders i.e. the highest degree of the derivative. They may be of the first order, second order, third order or more.
For example: dy/dx + y2 = 5x First Order Differential Equation.
d2y/dx2 + xy = sin(x) Second Order Differential Equation.
d3y/dx3 + x dy/dx + y = ex Third Order Differential Equation.
Second-Order Differential Equation Examples
In several real-world situations, First Order Differential Equation does not suffice properly and there is a need to implement Second Order Differential Equation. For a degree greater than one, there’s always a specific trick that is particular to the type of situation you are handling. For you to know how to handle the 2nd order Differential Equation, make sure you go through the concept completely as it will show you some examples and the technique involved in solving 2nd Order Differential Equation.
Difference Between 1st and 2nd Order Differential Equations
In the unknown y(x) Equation (1) is 1st order seeing that the highest derivative that seems in it is a 1st order derivative. Similarly, equation (2) is a 2nd order because also y appears. They are both linear, since y, y and y are not squared or cubed etc and their product is not appearing.
How to Solve Second Order Differential Equations?
In order to obtain the solution of the 2nd order differential equation, we will take into account the following two types of second-order differential equation.
For Homogeneous Second Order Differential Equation
The first type of equation you are going to handle are the ones like:
A d2y/dt2 + B dy/dt + C y = 0
Which can be written for both variables y and t. Here, A, B and C are constants which are generally coefficients that you don’t need to worry about.
Now the solution of Second Order Differential Equation starts by taking a guess which is a calculated guess. So we guess a solution to the equation of the form.
y = eλt, where λ is a constant.
Now we have to find λ for which a solution satisfies the second order Differential equation.
So, this implies dy/dt = λeλt, d2y/dt2 = λ2 eλt,
Thus, we can substitute this into our differential equation as:
a λ2eλt + b λeλt + c eλt = 0.
Now, we know that eλtis not equal to zero, then it effectively becomes a λ2 + b λ + c = 0.
This equation is generally called the Auxiliary Equation which you jump straight away. This is a quadratic equation and they are easily solvable. This means that we have to find λ for A, B and C that allow eλt that will satisfy the 2nd order differential equation. We will get two values of λwhich we assume as λ1 and λ2.
y = A eλ1t + B eλ2t.
what if λ1 = λ2, then the equation would be:
y = (A + Bt)eλ1t.
we should know that λmay be complex. If they are complex, then the solutions of λ1 and λ2 would be λ1 = r + is and λ2 = r – is.
Then our solution for y, using the relations between eλt and the trigonometric functions, can be written as:
y = ert (A cost+ B sint).
These are the steps you need to know. For homogeneous second order differential equation, simply move to the auxiliary equation, find the value of lambda λ, write the solution of y and then find the constants using initial conditions if required.
For Inhomogeneous Second Order Differential Equation
The extension to our problem could be when the Right Hand Side of our 2nd Order Differential Equation may be non-zero i.e.
A d2y/dt2 + B dy/dt + C y = f(t).
The solution here is quite simple where we can write the general solutions for y as y = yc + yp, where yc is known as the complementary function, and yp the particular integral.
This means that yc is the solution to the problem:
A d2yc/dt2 + B dyc/dt + C yc = 0,
And yp to the problem,
A d2yp/dt2 + B dyp/dt + C yp = f(t).
We should know that yc is found from the above case, so we just need to find the auxiliary equation. For yp, we again make a guess which depends on f. There are only a few guesses you need to know:
f(t) yp
eαt P eαt
αxn+ lower-order powers P xn + Q xn − 1 +... + Z
cosαt or sinαt P cosαt + Q sinαt
FAQs on Second Order Differential Equation Concepts and Methods
1. What is a second order differential equation?
A second order differential equation is a differential equation that involves the second derivative of an unknown function. In general form, it is written as F(x, y, y', y'') = 0 or explicitly as y'' = f(x, y, y'). It contains:
- The dependent variable y
- The first derivative y'
- The second derivative y''
2. What is the general form of a second order linear differential equation?
The general form of a second order linear differential equation is a(x)y'' + b(x)y' + c(x)y = r(x). Here:
- a(x), b(x), and c(x) are functions of x
- r(x) is the forcing (non-homogeneous) term
3. What is the difference between homogeneous and non-homogeneous second order differential equations?
The key difference is that a homogeneous equation has zero on the right-hand side, while a non-homogeneous equation does not. Specifically:
- Homogeneous: a(x)y'' + b(x)y' + c(x)y = 0
- Non-homogeneous: a(x)y'' + b(x)y' + c(x)y = r(x), where r(x) ≠ 0
4. How do you solve a second order linear homogeneous differential equation with constant coefficients?
To solve it, form and solve the characteristic equation associated with the differential equation. Steps:
- Given ay'' + by' + cy = 0
- Form the characteristic equation: ar² + br + c = 0
- Solve for roots r
5. What is the characteristic equation in second order differential equations?
The characteristic equation is the quadratic equation obtained by replacing y'' with r² and y' with r in a linear homogeneous equation with constant coefficients. For ay'' + by' + cy = 0, it is ar² + br + c = 0. Solving this quadratic gives the roots that determine the general solution.
6. What are the possible cases of roots in a second order differential equation?
There are three possible root cases for the characteristic equation ar² + br + c = 0. These are:
- Distinct real roots (r₁ ≠ r₂): y = C₁e^{r₁x} + C₂e^{r₂x}
- Repeated real root (r): y = (C₁ + C₂x)e^{rx}
- Complex roots (α ± βi): y = e^{αx}(C₁ cos βx + C₂ sin βx)
7. How do you find the general solution of a non-homogeneous second order differential equation?
The general solution is the sum of the complementary solution and a particular solution. It is written as y = y_c + y_p, where:
- y_c solves the homogeneous equation
- y_p is any particular solution of the non-homogeneous equation
8. Can you give an example of solving a second order differential equation?
Yes, for example, solve y'' − 3y' + 2y = 0. Steps:
- Characteristic equation: r² − 3r + 2 = 0
- Factor: (r − 1)(r − 2) = 0
- Roots: r = 1, 2
9. What are second order differential equations used for?
Second order differential equations are used to model systems involving acceleration or curvature. Common applications include:
- Simple harmonic motion and oscillations
- Mass-spring systems
- Electrical circuits (RLC circuits)
- Mechanical vibrations
10. What is the order and degree of a second order differential equation?
The order is the highest derivative present, and the degree is the power of the highest derivative after removing radicals and fractions. For example, in y'' + 3y' + y = 0:
- Order = 2 (highest derivative is y'')
- Degree = 1 (y'' is raised to power 1)
