Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Here is an equation: y equals x times d y by d x, plus the constant a, over d y by d x. Notice that the derivative is never squared anywhere. So you might say its degree is one. But the answer is two! So where does that two come from? This puzzle opens Part A of our unit problem bank, five problems at the JEE Main level. Problems like these come up often. So, let's begin.

0:33Part A: a 30-minute test, five problems

So, here is Part A: five mixed problems from the whole unit. The course suggests you attempt it as a thirty minute test. Problem one asks for the order and the degree of an equation. Okay? Problem two says y prime equals e to the power, two x plus y. The small dash after y is called a prime, and y prime means d y by d x. So which type is this one? Yes, it is separable, exactly! Because e to the power, two x plus y, is e to the power two x, times e to the power y; the dot means times. So the x part and the y part come apart. Right? Problems three and five are linear, so both of them need the integrating factor. Clear? And problem four says x y times y prime equals x squared plus y squared. Here x y, x squared and y squared all have degree two, so the equation is homogeneous. Got it? Today we take problems one and four, the easier one first, with one tool for each. Let's go.

1:43Tool 1: the order and the degree

Tool one comes from chapter one. The order of a differential equation is the order of the highest derivative present in it. Okay? And the degree is the power of that highest derivative, but only once the equation is written as a polynomial in the derivatives. Right? So before you read the degree, clear the radicals, and clear any derivative from a denominator. This is the key step. With me? Here is a quick test. The square root of y double prime equals y prime; the two dashes mean the second derivative, and the long sign over it is the square root. What is its degree? One half? Oh no, be careful here, because a degree is never a fraction. Square both sides first, and y double prime equals y prime squared. Now it is a polynomial, and y double prime, the highest derivative, has power one. So the degree is one. Got it? But y double prime equals sine of y prime can never be made a polynomial in y prime, because the derivative sits inside a sine. So that equation has no degree at all. Clear?

2:48Tool 2: homogeneous, so put y=vx

Tool two comes from chapter three. An equation is homogeneous when d y by d x equals capital F of y over x, so the right side depends only on the ratio, y over x. Okay? To test it, write the right side as a top over a bottom, and let small f stand for either one. Replace x by lambda x, and y by lambda y; lambda is the Greek letter on screen. If f comes out as lambda to the power n, times f, with the same n on top and bottom, then the lambdas cancel. Right? Then we put y equals v x, where v is a new function of x. Tell me, what is d y by d x? It is the product rule, exactly! d y by d x equals v, plus x times d v by d x. Many students forget the second term, so watch out. Clear? Now v and x separate. Integrate both sides, and at the end, put v back as y over x. Got it?

3:50Example 1: order and degree of y = xy' + a/y'

Now, our puzzle from the start, which is problem one of Part A, okay? Find the order and degree of y equals x times d y by d x, plus the constant a, over d y by d x. The slanted line at the bottom also means over. Pause the video and try it first. Done? Okay, let's check. The only derivative here is d y by d x, the first derivative, which we write as y prime. So the order is one. Right? Now the degree. y prime has power one in both terms, so the degree is one, isn't it? Wait, careful, because that is exactly the trap! Look at the last term, the constant a, over y prime. The derivative sits in a denominator, so this is not yet a polynomial in y prime. We must clear it first. Okay? So multiply every term by y prime. y times y prime equals x times y prime squared, plus the constant a. Right? Bring everything to one side: x times y prime squared, minus y times y prime, plus the constant a, equals zero. Now it is a polynomial in y prime, and its highest power is two. With me? So the order is one, and the degree is two. And that solves our puzzle! There was no square anywhere at the start, and yet the degree is two, because the square appears only after we clear the fraction. See?

5:24Example 2: solve xy y' = x squared + y squared with y = vx

Now problem four, which is a little harder. Solve x y times y prime equals x squared plus y squared. Which tool will you use here? Pause the video and try it first. Done? Okay, let's check. Divide by x y, and y prime equals x squared plus y squared, over x y. The top and the bottom both have degree two, so the equation is homogeneous, and tool two works. Right? Now put y equals v x. Tell me, what does the right side become? Every term has x squared, so it cancels from the top and the bottom. We get one plus v squared, over v, which is one over v, plus v. And the left side is v plus x times v prime, where v prime means d v by d x. Okay? Look, v sits on both sides, so it cancels! So x times v prime equals one over v. Multiply by v, and x times v times v prime equals one. So v d v equals d x over x, and the variables are separate. Clear? Integrate both sides. v squared over two equals log x, plus capital C, the arbitrary constant; on screen, l n means the natural log. Got it? Now put v back as y over x, and multiply by two x squared. Many students write y squared equals two x squared log x, plus capital C, and stop there. Is that right? Oh no, be careful here, because the constant gets multiplied by two x squared too. y squared over two x squared equals log x plus capital C, and the double arrow means so. So y squared equals two x squared, times the bracket, log x plus capital C, and that is the answer. One arbitrary constant, just right for a first order equation. Very good!

7:28Quick revision: clear denominators, multiply the constant

So, quick revision. In a mixed set, name the type first, and then pick the tool. For the degree, clear the radicals, and clear derivatives from denominators, before you read the power. For a homogeneous equation, put y equals v x, and when you clear the fractions at the end, multiply the constant too. Now go and try the practice problems on this page. You are ready for them. All the best!