Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Some quantity y is zero when x is zero, and d y by d x, plus two x times y, equals x. As x grows without end, does y grow forever, or settle down at some number? Hard to guess, right? By the end, you will find it in a few short lines. Questions like this come up often in JEE Main, so let's begin!

0:27How this chapter is built

So, where are the real exam questions? The JEE Main questions from twenty twenty one to twenty twenty six are at the end of Chapters one to five; the short dash means to. Okay? This chapter has original problems, in three mixed sets: eight, eight and six solved, with two practice problems each. Mixed means you spot the type yourself. Got it?

0:50The usual pattern: solve, then use y

Now, JEE Main questions here usually follow one pattern. Step one is an initial-value problem: an equation, plus a starting value of y. Okay? The equation is usually one of three types: separable, homogeneous or linear. So spotting the type comes first, right? Look at the figure, the slope field of y prime equals x minus y, where the dash, called prime, means the derivative. The short grey segments show slopes, and each curve is a solution, one curve for each value of the constant, capital C, okay? So how many arbitrary constants does the general solution of a first-order equation have? Yes, exactly, just one constant, capital C, for a first-order equation! So one starting condition fixes it: y of x naught equals y naught, where the small zero below is a subscript, and x naught, y naught is the starting point. In the figure, that picks exactly one curve. With me? Then step two: use the solution. The question asks for a value of y, an area, a limit or an extremum, that is, a maximum or a minimum.

2:01Type 1: separable

Clear? Type one, separable: d y by d x equals g of x times h of y, a function of x times a function of y. The double arrow means implies, and the tall S is the integral sign. Divide by h of y and integrate: the integral of d y over h of y equals the integral of g of x, d x, plus capital C, okay? But we divided by h of y. What could go wrong there? Yes, exactly, h of y could be zero! Dividing assumes h of y is not equal to zero, shown by the slashed equals sign. Right? So if h of y naught is zero, the constant y equals y naught is a solution that the division can throw away. Check it separately. Clear?

2:51Type 2: homogeneous, put y=vx

Type two, homogeneous: d y by d x equals f of x, y, over g of x, y. Scale x and y by lambda, the Greek letter on screen. If f of lambda x, lambda y equals lambda to the power n, times f of x, y, with the small n up high as the power, and g does the same with the same power n, the equation is homogeneous. Okay? Then the right side depends only on y over x, so d y by d x is capital F of y over x. So put y equals v x; by the product rule, d y by d x equals v, plus x times d v by d x. Right? So x times d v by d x equals capital F of v, minus v: separable, type one again! With me? But check first. Is x plus y plus one, over x minus y, homogeneous? No, it is not homogeneous, because the constant one is not multiplied by lambda; such an equation needs a shift of origin first. So check before you put y equals v x. Got it?

3:57Type 3: linear, use the integrating factor

Type three, linear: d y by d x, plus P of x times y, equals Q of x, with P and Q depending on x only. The coefficient of d y by d x must be one, so divide first if needed. Okay? The letters I and F stand for integrating factor: e to the power integral P d x, with no constant. Then y times the integrating factor equals the integral of Q times the integrating factor, d x, plus capital C; the dot means times. Right? Now, to get y, many students divide only the first term by the integrating factor, and stick plus capital C at the end. Is that right? No, careful, that is a classic trap! Capital C was added before dividing, so it is divided too: the term with capital C in y is capital C over the integrating factor, not capital C alone. Clear?

4:53Example 1: y' = 3x2 e^-y with y(0) = 0, find y(2)

Example one, from Set one, marked easy. y prime equals three x squared times e to the power minus y, and y of zero is zero. Find y of two. Pause the video, and try it first. Done? Okay, let's check. Which of our three types is it? Separable, yes, exactly: a function of x times a function of y. Multiply by e to the power y, and by d x: e to the power y, d y, equals three x squared, d x. Since e to the power minus y is never zero, no solution is lost. Clear? Integrate: e to the power y equals x cubed, plus capital C, okay? At x equals zero, y is zero, and e to the power zero is one. So one equals zero plus capital C, and capital C is one. Right? So e to the power y equals x cubed plus one. At x equals two, two cubed is eight, and eight plus one is nine, so y of two is the natural log of nine; l n means natural log. An initial-value problem, then a value: our pattern exactly!

6:08Example 2: limit of y as x tends to infinity, y' + 2xy = x

Now, remember our puzzle? Let's finish it. From Set three, marked hard: y prime, plus two x times y, equals x, with y of zero equal to zero. Find the limit of y as x tends to infinity: lim means limit, the arrow means tends to, and the sideways eight is infinity. Pause the video, and try it first. Done? Okay, let's check. It is linear, with P equal to two x. The integral of two x, d x, is x squared, so the integrating factor is e to the power x squared. With me? Here Q is x, so y times e to the power x squared equals the integral of x times e to the power x squared, d x. The derivative of x squared is two x, so this is one half e to the power x squared, plus capital C, okay? At x equals zero, y is zero and e to the power zero is one, so zero equals one half plus capital C, and capital C is minus one half. So y times e to the power x squared equals one half e to the power x squared, minus one half. Right? Now divide by e to the power x squared. Is y just one half minus one half, so zero for every x? No, careful, that is our trap again! The constant gets divided too. So y equals one half, minus one half times e to the power minus x squared, which is zero at x equals zero, as it should be. Got it? Now, as x tends to infinity, e to the power minus x squared tends to zero. So y tends to one half. It does not grow forever; it settles at one half, and that solves our puzzle!

7:54Quick revision: spot the type, solve, then use y

So, quick revision. The pattern: solve the initial-value problem, then use y. Spot the type: separable, homogeneous with y equals v x, or linear with the integrating factor. Watch the traps: lost solutions, a fake homogeneous equation, and capital C before dividing. Now open Set one and start solving. All the best!