The Integrating Factor

Transcript

0:00Introduction

Hello everyone! Take the equation d y by d x, plus y, equals e to the power x. Try to separate the variables, y on one side and x on the other. They refuse to come apart, right? Today we learn one clever multiplier, the integrating factor, that turns the whole left side into the derivative of a single product. This comes up often in JEE, so let's begin.

0:28One multiplier makes a product

So, here is our equation again. The dash on y is a prime, and y prime means d y by d x. So, y prime plus y equals e to the power x. Okay? Now a small question for you. What is the derivative of y times e to the power x? Use the product rule: y prime times e to the power x, plus y times e to the power x. Take e to the power x common, and look inside the bracket: y prime plus y, our left side! Beautiful, right? So, multiply both sides by e to the power x. The left side becomes this derivative, and the right side becomes e to the power two x. With me? Now integrate both sides: y times e to the power x equals one half e to the power two x, plus capital C, the constant. That multiplier is called the integrating factor. Clear?

1:27The linear equation and its integrating factor

d y by d x, plus P of x times y, equals Q of x, where P and Q depend on x only. This is the linear first order equation in standard form, which means the coefficient of d y by d x is one. If something multiplies d y by d x, divide by it first. Got it? In our example, P was one, and the multiplier was e to the power x. So can you guess the general rule? Yes, exactly, e to the power integral P d x, because the integral of one, d x, is x. The letters I and F stand for integrating factor, and the tall S is the integral sign. We add no constant here; one integrating factor is enough. Okay? Multiply by it, and the left side becomes d by d x of y times the integrating factor. The dot just means times. On the right, Q times the integrating factor. Right? Integrate both sides, and here is the key line: y times the integrating factor equals the integral of Q times the integrating factor, d x, plus capital C, the constant. This one line is the whole method, so write it down!

2:42The trap: add C before you divide

Now let's finish our first equation. y times e to the power x equals one half e to the power two x, plus capital C, okay? We only need to divide by e to the power x. Many students divide only the first term and stick plus C at the end, like this. Is that correct? No, no, careful, the constant was there before we divided! So capital C also gets divided, and one over e to the power x is e to the power minus x. So y equals one half e to the power x, plus capital C times e to the power minus x. Clear? So the rule is this: the constant term in y is capital C over the integrating factor, not capital C alone. With a starting value, this one slip gives a completely wrong answer.

3:30Topper's trick: integrating factors to know

Right? Now a trick toppers use: know the common integrating factors by heart. When P is n over x, the integral is n times the natural log of x, the l n on screen. And e to the power n times natural log x is x to the power n. The double arrow means gives. With me? Now you tell me. If P is tan x, what is the integrating factor? It is sec x, because the integral of tan x is the natural log of sec x. In the same way, cot x gives sin x. Okay? Two more: when P is one over the product x times natural log x, the integrating factor is the natural log of x. And when P is two x over one plus x squared, it is one plus x squared. Spot these, and you save time in the exam. Got it?

4:23When the equation is linear in x

One more case. Sometimes an equation is linear in x instead of y. Then flip the roles: d x by d y, plus P of y times x, equals Q of y. Right? Everything else is the same, with x and y swapped. The integrating factor is e to the power integral P d y, and x times it equals the integral of Q times the integrating factor, d y, plus capital C, the constant.

4:51Example 1: solve y' + 2y/x = x2 with I.F. = x2

Clear? Now, Example one. Solve y prime, plus two y over x, equals x squared. Pause the video and try it first. Done? Okay, let's check. It is already in standard form, with P equal to two over x. The integral of two over x is two log x, so the integrating factor is e to the power two log x, which is x squared. Straight from our list, with n equal to two. Nice, right? Multiply through by x squared. The left side becomes d by d x of x squared times y, and the right side is x to the power four. Okay? Integrate: x squared times y equals x to the power five over five, plus capital C, the constant. With me? Now we divide by x squared. Remember the trap. What happens to capital C here? Yes, exactly, it gets divided too! So y equals x cubed over five, plus capital C over x squared, which is capital C over the integrating factor. Clear?

5:59Example 2: y' + 2y = 4x with y(0) = 1

Now a harder one, Example two. Solve y prime plus two y equals four x, given y of zero equals one, which means y is one when x is zero. Pause the video and try it first. Done? Okay, let's check. Here P is two, so the integrating factor is e to the power two x. So d by d x of y times e to the power two x equals four x times e to the power two x. Okay? Integrate the right side by parts, with four x as the first function. You get two x minus one, times e to the power two x, plus capital C, the constant. Differentiate it to check, and four x times e to the power two x comes back. Right? Now the tempting move: divide by e to the power two x, and write y equals two x minus one, plus capital C, as on the badge. Can we do that? No, no, careful, that is our trap again! Capital C must be divided too. Safest is to use the starting value before dividing. Put x equal to zero and y equal to one; e to the power zero is one. So one equals minus one plus capital C, which gives capital C equals two. Clear? Now divide by e to the power two x. y equals two x minus one, plus two times e to the power minus two x. At x equal to zero, we get minus one plus two, which is one, perfect. See, the constant sits with e to the power minus two x, as the rule said.

7:42Quick revision: standard form, I.F., add C before dividing

So, quick revision. First, standard form, and the integrating factor is e to the power integral P d x. Second, y times the integrating factor equals the integral of Q times the integrating factor, d x, and add capital C before you divide. Third, if the equation is linear in x, swap the roles of x and y. Now go and try the practice problems on this page. All the best!