Exact Differentials

Transcript

0:00Introduction

Hello everyone! Here is an equation for you: x d y, plus y d x, equals x squared d x. Try to separate the variables, with all the y terms on one side and all the x terms on the other. The left side mixes x and y together, so they refuse to come apart, right? But look at it again with fresh eyes. That whole left side is the change in one single quantity, x times y. Today we learn to spot such combinations, and a simple test for exact equations. This is a handy tool for JEE problems, so let's begin.

0:37Spot one differential: the product xy

So, let us start with the product rule. The upright d you see stands for a differential, a tiny change. If x and y both change a little, what is the change in x times y, that is, d of x y? Yes, exactly, it is x d y, plus y d x! That is just the product rule, written with differentials. Okay? Now look at our equation from the start. Its left side, x d y plus y d x, is exactly d of x y! The double arrow means gives, so d of x y equals x squared d x. Right? Now integrate both sides: x y equals x cubed over three, plus capital C, the constant. We needed just two lines! That is the whole idea: spot a group of terms that is the differential of one function, then integrate. Clear?

1:32Integrable combinations to know

Now, there is a short list of such combinations to know, and the first ones come from the quotient rule. So tell me, what is d of y over x? It is x d y minus y d x, all over x squared. The horizontal bar is division, so read it as over. So when you see x d y minus y d x, think of dividing by x squared. With me? Now let us flip the fraction. What is d of x over y? Which term comes first this time? This time y d x comes first: y d x minus x d y, over y squared, is d of x over y. Be careful here, because the order in the numerator flips; swap it, and your whole answer changes sign. Got it? Now keep the same numerator, x d y minus y d x, but divide by x squared plus y squared. That is d of tan inverse of y over x. The little minus one on tan means inverse, not one over tan. Okay? Next, x d y plus y d x, over x times y, is d of the natural log of x y. The letters l n on screen stand for the natural log. Right? And the last one: x d x plus y d y, over the square root of x squared plus y squared, is d of that same square root. So here is the habit: read the numerator first, and it tells you which line to try.

3:04Exact equations: the test and the solution

Clear? Alright, not every equation hands us a ready combination. So write it as M d x, plus N d y, equals zero, where capital M and capital N are functions of both x and y. Okay? Now, when is this left side exactly d F, the differential of one single function F? How can we tell, without guessing? Here is the test. The curly d means a partial derivative. Differentiate M with respect to y, treating x as a constant, and N with respect to x, treating y as a constant. If partial M by partial y equals partial N by partial x, the equation is exact. Clear? Then how do we find F? First, integrate M with respect to x, holding y constant. The tall S is the integral sign. With me? Next, look at N, and pick only the terms that have no x in them. Integrate those with respect to y, and add them on. Right? And the solution is F of x, y equals capital C, the constant. Test, integrate, add, and you are done. Got it?

4:18Example 1: x dy - y dx = x2 e^x dx using d(y/x)

Now, Example one. Solve x d y, minus y d x, equals x squared times e to the power x, d x. Pause the video and try it first. Have you tried it? Okay, let's check. The left side is x d y minus y d x, just like the first line of our list. So what should we divide both sides by? Yes, by x squared, exactly! Dividing by x squared assumes x is not zero, so the screen says x not equal to zero, the equals sign with a slash through it. On the right, the x squared cancels, and only e to the power x, d x is left. Okay? Now the left side is d of y over x, so it equals e to the power x, d x. Integrate both sides, and y over x equals e to the power x, plus capital C, the constant. Right? Finally, multiply by x. So y equals x times the bracket, e to the power x plus capital C, the constant. Keep the bracket, because the constant gets multiplied by x too. Clear?

5:31Example 2: e^x sin y dx + (e^x cos y + 2y) dy = 0

Now a harder one, Example two. Solve e to the power x times sine y, d x, plus the bracket, e to the power x times cos y, plus two y, times d y, equals zero. Pause the video and try it first. Have you done it? Okay, let's check. No combination jumps out here, so we use the exact test. M is e to the power x times sine y, and N is e to the power x times cos y, plus two y. With me? Now you do the test. Find partial M by partial y, and partial N by partial x. Are they equal? Yes, exactly, they are equal! In M, e to the power x acts as a constant, and sine y gives cos y. In N, the first term gives e to the power x times cos y, and two y has no x, so it gives zero. Both are e to the power x times cos y, so the equation is exact. Okay? Now integrate M with respect to x, holding y constant, and we get e to the power x times sine y. Right? Now the tempting move: integrate all of N with respect to y, and add it on. Should we do that? No, wait, because that is exactly the trap. The integral of all of N in y is e to the power x times sine y, plus y squared. But e to the power x times sine y is already in F, so it would be counted twice. Got it? So add only the term of N without x, which is two y. Its integral with respect to y is y squared. Clear? So the solution is e to the power x times sine y, plus y squared, equals capital C, the constant. Differentiate it to check, and you get back exactly our M and N, perfect! Very good, you have solved an exact equation!

7:30Quick revision: spot combinations, test exactness, F = C

So, quick revision. First, spot a combination, like x d y plus y d x, which is d of x y, and the others from our list. Second, the exact test: partial M by partial y equals partial N by partial x. Third, integrate M in x, add only the terms of N without x, integrated in y, and set F equal to capital C, the constant. Now go and try the practice problems on this page. All the best!