Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Here is a quick puzzle. Take the equation y prime plus y equals y squared, where y at zero is one half. Find y at the natural log of three. That y squared looks like trouble, right? Is this equation even linear? Not yet, but one small substitution will make it linear, and the answer comes out as a neat little fraction. In a mixed set, nobody tells you the method, so spotting the type is half the work. So let's practise exactly that in Set two. Let's go.

0:34The pattern of a mixed problem

So, first look at how these questions are built. Many JEE Main questions on differential equations follow one pattern. Step one: spot the type, separable, homogeneous, or linear. Okay? Step two: solve it, and use the initial condition, the value of y given at one point, to find capital C, the constant. Right? And step three: finish the question, by finding a value, an area, a limit, or an extremum, that is, a maximum or a minimum. Clear?

1:09Tool 1: the linear form and the I.F.

Now our first tool, the linear equation: d y by d x, plus P of x times y, equals Q of x. The dash on y is a prime, so y prime means d y by d x, and in the standard form its coefficient must be one. Okay? I F on screen means the integrating factor. The tall S is the integral sign, so the integrating factor is e to the power integral P d x, with no constant inside. Right? Multiply by it, and the left side becomes the derivative of y times the integrating factor. The small dot means times. So y times the I F equals the integral of Q times the I F, d x, plus capital C, okay? Now a quick one for you. If P is cot x, what is the integrating factor? The integral of cot x is the natural log of sine x, taking sine x positive, and e to the power log of sine x is just sine x. The double arrow means gives, so P equals cot x gives an integrating factor of sine x. Got it? And here is the trap: add capital C first, and only then divide by the integrating factor. So the term with capital C in y is capital C over the I F, not capital C alone. Clear?

2:30Tool 2: Bernoulli becomes linear

Now the second tool, for equations that are almost linear. Bernoulli's equation: y prime, plus P of x times y, equals Q of x times y to the power n; the small raised n is the power. Here n is not zero and not one, or the equation would already be linear; the equals sign with a slash means not equal to. Okay? So, divide by y to the power n, which assumes y is not zero, as the round bracket says. Then put v equals y to the power one minus n. Right? By the chain rule, the equation becomes d v by d x, plus the bracket one minus n, times P times v, equals the bracket one minus n, times Q, which is linear in v. So tool one finishes it. And careful, the factor one minus n multiplies the right side too. With me? For n equals two, one minus n is minus one, so v is y to the power minus one, that is, one over y. We need this soon. Got it?

3:32Example 1: y' + y cot x = 2 cos x, find y(pi/6)

Now, Example one, the easier one. y prime, plus y times cot x, equals two cos x. And y at pi over two is one; pi is the Greek letter, and the round bracket holds the value of x. Find y at pi over six. Pause the video, and try it first. Done? Okay, let's check. Step one of our pattern: which type is this equation? Yes, exactly, it is linear, and already in standard form, since y prime has coefficient one. P is cot x, and Q is two cos x. So from tool one, the integrating factor is sine x. Okay? Multiply through by sine x. The left side becomes the derivative of y times sine x; the dash outside the bracket means differentiate everything inside. The right side is two sine x cos x, which is sine two x. Right? Now integrate: the integral of sine two x, d x, is sine squared x, because the derivative of sine squared x is two sine x cos x. And we add capital C right here, before dividing. Clear? Now step two of the pattern: we know y at pi over two is one. So what is capital C? Put x equals pi over two. Sine of pi over two is one, so the left side is one times one. The right side is one plus capital C, so capital C is zero. Got it? So y times sine x equals sine squared x, and dividing by sine x gives y equals sine x, really simple, right? Finally, y at pi over six is sine of pi over six, which is one half. That's it, the answer is one half!

5:22Example 2: Bernoulli y' + y = y squared, put v = 1/y

Now, remember our puzzle from the start? Let's solve it. Example two: y prime plus y equals y squared, and y at zero is one half. Find y at l n three, the natural log of three. Pause the video, and try it first. Done? Okay, let's check. Step one again: is this equation linear? No, the y squared spoils it. But it is Bernoulli, with P and Q both equal to one, and n equal to two. So divide by y squared, assuming y is not zero, and put v equals one over y. y equals zero does satisfy the equation, a lost solution, but it fails the condition y at zero equals one half. Okay? After dividing, the first term is y prime over y squared. Is that just v prime, so that the equation is v prime plus v equals one? No, careful here, because by the chain rule, v prime is minus y prime over y squared, so that first term is minus v prime. We get minus v prime plus v equals one, so v prime minus v equals minus one. That minus one on the right is the factor one minus n at work. Clear? Now it is linear in v, with P equal to minus one. So the integrating factor is e to the power minus x, and the derivative of v times e to the power minus x equals minus e to the power minus x. With me? Integrate, adding capital C first: v times e to the power minus x equals e to the power minus x, plus capital C, right? Multiply by e to the power x, and v equals one plus capital C times e to the power x. Now the initial condition: y at zero is one half, so v at zero is two. Two equals one plus capital C, so capital C is one. Got it? So v is one plus e to the power x, and y is one over that bracket. At x equals l n three, e to the power l n three is three, so y is one over the bracket, one plus three, which is one quarter. And there is our neat little fraction, puzzle solved!

7:40Quick revision: linear I.F., Bernoulli swap, spot the type

So, quick revision. Linear: write the standard form, take the integrating factor e to the power integral P d x, and add capital C first. Bernoulli: divide by y to the power n, put v equals y to the power one minus n, and watch the sign of v prime. And the pattern: spot the type, find capital C, then finish the question. Now go and try the practice problems on this page. You will do well, I promise. All the best!