JEE Advanced Questions: Homogeneous Equations

Transcript

0:00Introduction

Hello everyone! Here is a question from JEE Advanced, twenty sixteen. x squared, plus x y, plus four x, plus two y, plus four, all times d y by d x, minus y squared, equals zero. Squares, single powers and a plain constant four, all mixed up. Does that look homogeneous? Not at all, right? And yet, it is a homogeneous equation in disguise. Today we pull off that disguise and settle every option. Let's begin.

0:32Homogeneous equations: a quick recap

So, first a quick recap from this chapter. The Greek letter you see is lambda, just a scaling number. We call f homogeneous of degree n if f of lambda x, lambda y equals lambda to the power n, times f of x, y. Okay? Now, if d y by d x equals f over g, with f and g of the same degree, the right side depends only on y over x. The little dash on y is a prime: y prime means d y by d x, and it equals capital F of y over x. Right? So we put y equals v x. By the product rule, y prime equals v, plus x times d v by d x, and the equation becomes separable. Clear? But first, always check homogeneity. x plus y plus one, over x minus y, is not homogeneous, because the constant one spoils it. And our JEE Advanced bracket has a constant four in it. With me?

1:34Pull off the disguise

Now, back to our question, for x greater than zero. The constant four spoils homogeneity, so y equals v x will not work straight away, okay? So let me ask you this: can you factorise that bracket? Here is a hint: look for x plus two. Yes, exactly! It is x plus two, times the bracket x plus two plus y. Multiply it out and check: you get back the whole bracket. And now x appears only inside x plus two. Got it? So give it a new name: capital X equals x plus two, a shift of origin. Since two is a constant, d capital X equals d x, so d y by d x is the same as d y by d capital X, okay? Now the equation reads: capital X, times capital X plus y, times d y by d capital X, equals y squared. Degree two on both sides. The disguise is gone, and it is homogeneous! Clear?

2:36Solve with y = vX, curve through (1, 3)

So put y equals v times capital X, where v prime means d v by d capital X, okay? The left side becomes v, plus capital X times v prime. On the right, divide the top and the bottom by capital X squared, and you get v squared, over one plus v. Now move that v to the right side. What is left after you simplify? Over the common denominator, the top is v squared, minus v times one plus v. The v squared cancels, leaving minus v. So one plus v, times capital X times v prime, equals minus v. Right? Now separate. We divide by v, so we assume v is not equal to zero, that is the crossed equals sign at the end, and we split one plus v, over v. The tall S shape is the integral sign: the integral of one over v, plus one, d v, equals minus the integral of d capital X over capital X, okay? The letters l n mean the natural log. So log v, plus v, equals minus log capital X, plus small c. Here capital X is positive, and y stays positive on our curve through one comma three, so v is positive too, and we need no modulus. Bring log capital X across: log v plus log capital X is the log of v times capital X, which is just log y! So log y, plus y over capital X, equals c. Beautiful, isn't it? Now, the curve passes through the point one comma three. So what is c? At x equals one, capital X is three, and y is three. So c is log three, plus three over three, which is log three, plus one. If you got that, very well done! Put capital X back as x plus two, and here is our solution curve: log y, plus y over the bracket x plus two, equals log three, plus one. Always come back to the original variable. Got it?

4:43Example 1: does the curve meet the line x = 1 again

Now, Example one. For this curve through one comma three, find y where it meets the line x equals one again, if at all. Pause the video and try it first. Done? Okay, let's check. At x equals one, x plus two is three. So log y, plus y over three, equals log three, plus one. Okay? Now, y equals three works, of course, because that is our starting point. But could there be a second value of y? Look at the left side. Log y grows as y grows, and y over three grows too, so their sum is increasing for every positive y. Right? And an increasing function takes each value only once, so y equals three is the only solution. So the curve meets the line x equals one only at one comma three, and never again. Neat, isn't it?

5:43Example 2: meeting y = x + 2, (x+2) squared, (x+3) squared

Now, Example two: the full question from the start, with x greater than zero and the curve through one comma three. Option A says it meets y equals x plus two at exactly one point, and option B says exactly two points. Option C says it meets y equals x plus two, whole squared. Option D says it does not meet y equals x plus three, whole squared. More than one can be true. Pause the video and try it first. Done? Okay, let's check. Our curve is log y, plus y over capital X, equals log three, plus one. And since x is greater than zero, capital X is greater than two. Do not forget that, okay? Now take options A and B, on the line y equals x plus two. Here y is exactly capital X, so what is y over capital X? It is one, of course! So log y, plus one, equals log three, plus one, which gives y equals three, and then x equals one. Exactly one meeting point, one comma three. Clear? So option A is true, and option B is false, because the equation left us only one value of y. With me? Now option C, where y is capital X squared. What does the left side of our curve become? Log of capital X squared is two log capital X, and capital X squared over capital X is capital X, so we get two log capital X, plus capital X, which increases. At capital X equals two it is already log four, plus two, bigger than log three, plus one. So for capital X greater than two there is no meeting point, and option C is false. Oh no, careful with this trap: if you forget that x is greater than zero, this increasing function does reach log three, plus one, somewhere below capital X equals two, and you would wrongly mark option C as true. Got it? Last, option D: y equals x plus three, whole squared. This y is even bigger than capital X squared, so log y and y over capital X both get even larger, and again there is no meeting point. Oh no, careful here: option D is worded with a not, so no meeting point means option D is true! So the answer is options A and D, together. One factorisation, one shift, one substitution, and the question is done. That's it!

8:23Quick revision: factorise, shift, y = vX, meeting points

So, quick revision. When a bracket looks ugly, factorise it, and look for a shift like capital X equals x plus two. Once it is homogeneous in capital X and y, put y equals v times capital X, separate, and come back to x. And to count meeting points, substitute, then use increasing functions. Now try the practice problem on this page. It uses the very same idea. All the best!