Using D-Operators As Objects To Solve An ODE .

Now that we’ve climbed the Olympus of abstractness when we explored what Objects and Operators are, I believe you are ready to put what you know to good use as we now focus our attention on the D-Operator method, an algorithm for getting the particular solution of an Ordinary Differential Equation. Before we get into it, we need to be conversant with some terminologies.

Ordinary Differential Equations(O.D.E)

Equations like 2x+3=y and x²+x=9 are examples of what is called an algebraic equation which basically describes a relationship between given variables. Differential equations on the other hand, is an equation relating an unknown function y(t),some of the derivatives of y(t), and the variable x. For example;

dy/dx + 2x=y

is an example of what’s called an Ordinary Differential Equation(O.D.E). This is distinguished from partial differential equation where the dependent variable y=f(x1,x2,x3,…,xn) is a function of more than one variable. Our focus here will be on O.D.E’s.

Solution Of An O.D.E

The solution to an O.D.E is a function y(t)that makes the equation true. for example, y=e^x is a solution to d²y/dx² -2dy/dx + y = 0. To see why observe that d²y/dx²=dy/dx=y=e^x so (e^x)-2(e^x)+(e^x)=0 and thus y=e^x is a solution but not the solution since y=xe^x is also a solution.

observe that d(xe^x)/dx=xe^x + e^x and d²(xe^x)/dx²=xe^x + 2e^x so

d²y/dx²-2dy/dx + y = (xe^x + 2e^x)- 2(xe^x +e^x)+xe^x=0.

It turns out that the general solution is actually just a linear combination of these two,ie y= Ae^x + Bxe^x.

This type of O.D.E where each derivative and dependent variable is of degree 1 and each have constants as coefficients is called a constant coefficient differential equation of order n(in the example n=2,from the highest derivative). Specifically this is a homogeneous differential equation where the right hand side is 0.

The particular solution is simply a solution to a non homogeneous ODE . For example, d²y/dx² -2dy/dx + y = x²+1.

D-Operators

There are different notations for the derivative of y, dy/dx and y’ are some candidates but for the sake of our discourse we shall focus on the notation Dy where D^n(y) stands for the nth derivative of y,e.g D⁶y is the 6th derivative of y.

Observe that we can rewrite ODE from our last example using this notation as:

D²y — 2Dy + y=x²+1.

Recall that from our discourse on Operators we drew a rigid line between operators and objects but for now we shall blur it a bit. Now I know your alarm bells are going off but calm down a bit. I’m not going to perform a blunder like saying sin/cos=tan but I am going to treat the differential operator(D-Operator) as though it is an object and apply the distributive law and factor out the y,

(D²-2D+1)y=x²+1.

while this might look illegal it actually reverts to the original ODE when you open up the brackets so yeah, it looks weird but it is actually correct. Observe that we can simplify further using rules of algebra on D to get

(D-1)²y=x²+1

Okay what is our aim after performing such abominable acts? To find y, and to find y we make it subject of the formula;

y = (D-1)-²(x²+1).______(1)

Observe that 1+D+D²+D³+…=1/(1-D)=(D-1)-¹ from geometric series. We can manipulate this result to look like (D-1)-² by multiplying through by -1 and squaring both sides to have that

(D-1)-²=(1+D+D²+D³+…)².

Inputting this result in the (1) gives:

y=(1+D+D²+D³+…)²(x²+1)._______(2)

Okay if you are with me up till this stage give yourself a pat on the back. You’ll need to stay with me for one more act of heresy before the punchline.

Why did I make the reduction from (1) to (2)? Remember when I said we blurred the line between operators and objects? Well the line is still there so is the nonsense notion of dividing operators doesn’t make sense. To make it make sense we need it to be eliminate it from being a divisor which is what we did to get (2).

Now from (2), we open brackets using the traditional approach;

y=((1+D+D²+D³+…)+(D+D²+D³+…)+(D²+D³+…)+(D³+D⁴+…))(x²+1)

Notice that this looks like a complete mess but our saving grace is that D^(n)(x²+1)=0 for n greater than or equal to 3 since differentiating x²+1 three times already results in 0.

Thus, y = (x²+1)+D(x²+1)+D²(x²+1)+…+D(x²+1)+D²(x²+1)+D³(x²+1)+…+D²(x²+1)+D³(x²+1)+…+D³(x²+1)+D⁴(x²+1)+…

y=x²+1+2x+2+0+0+…+2x+2+0+0+0+…+2+0+0+0+0+…

y=x²+4x+6

and we have found a particular solution!

Hope you enjoyed the adventure. Happy mathing!