Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Imagine a population that starts at one hundred and grows by this rule: d P by d t equals P times, one minus P over one thousand. When does it reach five hundred? Looks like a long calculation, right? By the end of this video, you will do it in a few clean lines. This chapter is the problem bank for the whole unit: problems from across the whole unit, all mixed together, the way a real exam paper mixes them. So let us see how to use it well. Let's begin!

0:36How to use this chapter

So, let's open this chapter. Part A has five solved problems at JEE Main level: four easy, and one medium. Okay? Now tell me, when should you attempt Part A? Straight after Chapter one? No, only after Chapters one to six; the short dash means to. And attempt it as a thirty minute test, with a clock running, right? Now a quick one for you: five problems in thirty minutes, so how much time does each one get? Thirty divided by five is six; that is the division sign on screen. So, six minutes a problem. Clear? Then Part B: four longer problems at JEE Advanced level, all marked hard. Our second example comes from there. Got it? And Part C: five concept checks, on the ideas that are often tested in statement form. With me?

1:33Topper's trick: name the type first

Now a trick that toppers use. In a mixed test, nobody tells you the chapter, so name the type first. Look at the badge on screen: y prime equals e to the power, two x plus y; y prime, with one small dash, means d y by d x. Which type is it? It is separable, yes, exactly! e to the power, two x plus y, is e to the power two x, times e to the power y. So e to the power minus y, d y, equals e to the power two x, d x. Right? Next, y prime plus y equals one: the linear form, y prime plus P y equals Q, with P and Q both one. I F on screen means the integrating factor, e to the power integral P d x; the long S is the integral sign. Here it is e to the power x, with no constant. Okay? And the third: x times y times y prime, equals x squared plus y squared. Is it separable, linear, or something else? It is homogeneous! Divide by x times y, for x and y not zero: y prime equals x squared plus y squared, over x times y. Top and bottom both have degree two, so we put y equals v x. Clear?

2:58Part C: a solution, or the general solution?

Now try a check from Part C, the concept checks. Is y equals x squared a solution of x times y prime equals two y? Put it in. y prime is two x, so x times two x is two x squared, exactly two y. The dot means times. So yes, it is a solution. Got it? So then it must be the general solution, right? Careful here, because the answer is no. The general solution is y equals capital C times x squared, where capital C is any constant. Our y equals x squared is just the member with capital C equal to one. With me? Why one constant? Because the general solution of a first order equation has exactly one arbitrary constant. In general, order n means n arbitrary constants.

3:52Example 1: order and degree of y = xy' + a/y'

Okay? Now our first example, from Part A, marked easy. Find the order and degree of this: y equals x times d y by d x, plus a second term, the constant a, over d y by d x. The slash in that bottom part just writes d y by d x on one line. Pause the video and try it first. Done? Okay, let's check. The only derivative here is d y by d x, so the order is one. Easy, right? Now the degree. In the first term, d y by d x has power one. So the degree is one, isn't it? Oh no, careful, look at the second term, where d y by d x sits in the denominator! The degree is defined only for a polynomial in the derivatives, so clear the fraction first. Multiply both sides by y prime, which is not zero, since it was already in a denominator. We get y times y prime equals x times y prime squared, plus the constant a. Okay? Now bring everything to one side: x times y prime squared, minus y times y prime, plus the constant a, equals zero. The highest power of y prime is two. So the order is one, and the degree is two! The fraction was hiding the real degree. Clear?

5:19Example 2: logistic dP/dt = P(1 - P/1000), time when P = 500

Now, remember our puzzle? It is a Part B problem, marked hard. d P by d t equals P times, one minus P over one thousand, and P of zero, the population at time zero, is one hundred. Find P as a function of t, and the time at which P equals five hundred. Pause the video and try it first. Done? Okay, let's check. One minus P over one thousand is one thousand minus P, over one thousand. So the variables separate: one thousand d P, over P times, one thousand minus P, equals d t. The arrow means implies. Dividing by P times, one thousand minus P, assumes P is not zero and not one thousand; the slashed equals sign means not equal. Those constant solutions are lost, but our P starts at one hundred, so neither is ours. Right? Now split it by partial fractions: d P over P, plus d P over, one thousand minus P, equals d t. Add them back, and the top is one thousand again. With me? Now we integrate both sides. But tell me, what is the integral of d P over, one thousand minus P? It is minus the natural log of, one thousand minus P, and that minus sign is the common slip! On screen, l n means natural log. So natural log of P, minus natural log of, one thousand minus P, equals t plus capital C; with P between zero and one thousand, no modulus is needed. Got it? Combine the logs and raise e to the power of both sides: P over, one thousand minus P, equals capital A times e to the power t, where capital A is e to the power capital C, a constant. At t equal to zero, P is one hundred, so capital A is one hundred over nine hundred, which is one ninth. Okay? So P over, one thousand minus P, equals one ninth of e to the power t. Cross multiply, collect the P terms, and divide top and bottom by e to the power t. P equals one thousand over, one plus nine times e to the power minus t. Clear? Now the last part: when is P equal to five hundred? The ratio form, left of the arrow, is quicker. When P is five hundred, one thousand minus P is also five hundred, so the ratio is one. So one ninth of e to the power t equals one, e to the power t equals nine, and t is the natural log of nine. That is our puzzle, solved in a few clean lines.

8:05Quick revision: test plan, degree rule, lost solutions

Right? So, quick revision. Part A: a thirty minute test after Chapters one to six; name the type first. For the degree, clear radicals and fractions of the derivatives first. And when you divide by a factor h of y, its zeros can be lost constant solutions, so check them. Now start your clock, and try Part A and the practice problems on these pages. All the best!