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.
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.
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.
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:
eαt P eαt
αxn+ lower-order powers P xn + Q xn − 1 +... + Z
cosαt or sinαt P cosαt + Q sinαt
Q1. How do we Find a Solution to the 2nd Order Differential Equation?
Answer: For finding a solution of second-order differential equation constants in yp, we have to just differentiate twice and put it into the Second Order Differential Equation. We have yc and yp, so we can easily find out y and use the initial conditions to find constants in yc.
Take this as an example:
So, we can find the yc¬using the auxiliary equation:
After this, we have:
The Right Hand Side becomes a polynomial, therefore:
Even though there is no ‘t’ in the RHS, we must use it in yp. This means:
Putting the values in the differential equation, we have
From the equation, we get P = 1. Therefore by equating we get Q = -4/5.
By equating t0, we get:
After all this, we get our solution for y as:
As now you know the solution of Second Order Differential Equation, just start your practice and continue progressing.
Q2. Why does a 2nd Order Differential Equation have Two Solutions?
Answers: A second-order differential equation in the linear form needs two linearly independent solutions such that it obtains a solution for any initial condition, say, y(0) = a, y′(0) = b for arbitrary 'a', 'b'. In terms of mechanical perspective, the position and the velocity can be determined independently.