Bernoulli Equations and Other Substitutions

Transcript

0:00Introduction

Hello everyone! Here is an equation for you: d y by d x, plus y over x, equals x squared times y squared. It looks almost linear, right? But that y squared on the right spoils it. Here is a little puzzle: how do we make that y squared go away? One clever substitution will make it linear again. Questions like this do appear in JEE, so let's begin.

0:28The Bernoulli equation

So, here is today's family, the Bernoulli equation. The dash on y is a prime, and y prime means d y by d x. On the left, y prime plus P of x times y, just like a linear equation. On the right, Q of x times y to the power n; the small raised n is the power. Okay? Now, let me ask you this: what happens when n is zero, or when n is one? If n is zero, y to the power zero is one, so we get the linear equation from the last lesson. If n is one, move Q times y to the left, and it is linear again. Nothing new, right? So Bernoulli needs n not equal to zero and not equal to one; the equals sign with a slash means not equal to. For every other n, like two or three, the y to the power n spoils linearity. Clear?

1:22Substitution v = y to the power 1 - n makes it linear

First, divide by y to the power n, assuming y is not zero. The first term becomes y to the power minus n, times y prime. The middle term becomes P times y to the power one minus n, and the right side is just Q of x. With me? So, here is my question: what is the derivative of y to the power one minus n, with respect to x? Yes, exactly, it is the chain rule. Put v equals y to the power one minus n. Then v prime equals one minus n, times y to the power minus n, times y prime. And see, y to the power minus n times y prime is exactly our first term! Okay? So, from line two, our first term is v prime over one minus n. Multiply everything by one minus n: d v by d x, which is just v prime, plus one minus n, times P, times v, equals one minus n, times Q of x. Here is the trap: one minus n multiplies the right side too, and students often forget it. Got it? Now this equation is linear in v, so the integrating factor finishes the job. At the end, put back v equals y to the power one minus n. Clear? Remember, for n equals two, v is one over y, and for n equals three, v is one over y squared.

2:45Other substitutions: t = f(y) such as cos y, tan y, ln y

Right? The same idea works beyond Bernoulli. Here f of y is some function of y, and f prime of y is its derivative. So we have f prime of y, times y prime, plus P of x times f of y, equals Q of x. Okay? Let me ask you again: what is the derivative of f of y, with respect to x? It is the chain rule again: f prime of y, times y prime, which is exactly the first term! So put t equals f of y, and t prime equals f prime of y, times y prime. With me? Now the equation becomes t prime, plus P times t, equals Q, which is linear in t. Right? The common choices for t are e to the power y, cos y, tan y, the natural log of y, written l n y, and y squared. Okay? And here is a topper's trick. Look for a function of y and its derivative times y prime, together in the equation. Spot that pair, and you have your substitution. Got it?

3:49Example 1: dy/dx + y/x = x squared y squared, put v = 1/y

Now, remember our equation from the start? Let's finish it. Example one: solve y prime, plus y over x, equals x squared times y squared. Pause the video, and try it first. Done? Okay, let's check. First, what is n here, and what should v be? The power of y on the right is two, so n is two, and one minus n is minus one. So divide by y squared, put v equals one over y, and v prime equals minus y prime over y squared. Okay? After dividing, we get y prime over y squared, plus one over x y, equals x squared. The double arrow means this leads to minus v prime, plus v over x, equals x squared. Right? Multiply by minus one: v prime, minus v over x, equals minus x squared. That minus on the right is the factor one minus n at work. Clear? Here P is minus one over x. I F on screen means the integrating factor: e to the power integral P d x, which is e to the power minus natural log of x, or one over x. Multiply through, and the left side becomes the derivative of v over x, which equals minus x. The dash outside the bracket means differentiate everything inside. With me? Integrate: v over x equals minus x squared over two, plus capital C, okay? Add the constant now, before multiplying by x; adding it after is a very common mistake. Multiply by x, and put v back as one over y. So, one over y equals capital C times x, minus x cubed over two. That answers our puzzle: putting v equals one over y made the y squared go away. Nice, right?

5:46Example 2: sin y dy/dx + cos y = x, put t = cos y

Now a slightly harder one. Example two: solve sine y times y prime, plus cos y, equals x. This is not Bernoulli, so which substitution will you use? Pause the video and try it first. Done? Okay, let's check. We spot the pair, cos y and sine y times y prime, so we put t equals cos y. Now, is t prime just sine y times y prime? Think carefully. No, no, careful, the derivative of cos y is minus sine y. So t prime equals minus sine y, times y prime. Miss this minus, and everything after it goes wrong. Got it? So the equation becomes minus t prime, plus t, equals x. Multiply by minus one: t prime minus t equals minus x, which is linear in t. Clear? Here P is minus one, so the integrating factor is e to the power minus x. Multiply through: the derivative of t times e to the power minus x equals minus x times e to the power minus x. With me? Now, how do we integrate minus x times e to the power minus x? Yes, exactly, integration by parts. The tall S shape is the integral sign. So t times e to the power minus x equals the integral of minus x times e to the power minus x, d x, which is x plus one, times e to the power minus x, plus capital C, okay? Finally, multiply by e to the power x, and put t back as cos y. So, cos y equals x plus one, plus capital C times e to the power x. The minus sign was the only real danger, see?

7:34Quick revision: v = y to the power 1 - n, or t = f(y)

So, quick revision. For Bernoulli, with n not zero or one, divide by y to the power n, and put v equals y to the power one minus n. Then v prime, plus one minus n times P times v, equals one minus n, times Q, which is linear, so use the integrating factor. For other forms, spot f of y with f prime of y times y prime, and put t equals f of y. Now go and try the practice problems on this page. All the best!