Equations Reducible to Homogeneous Form

Transcript

0:00Introduction

Hello everyone! Here is a friendly looking equation. d y by d x equals x plus two y minus three, over two x plus y minus three. Last lesson, we solved homogeneous equations by putting y equals v x. So, does y equals v x work here? Not really, because those two minus threes spoil everything, right? Today we make them disappear, just by moving the origin to one clever point. This type does appear in JEE questions, so let's begin.

0:34When the constants spoil it

So, here is today's family. The dash on y is a prime, and y prime means d y by d x. On top we have small a times x, plus b times y, plus c. In the bottom, the dashes on small a, b and c just mark three new constants: a prime, b prime and c prime. Okay? Now tell me, what stops this equation from being homogeneous? Yes, exactly, the constants c and c prime! Without them, every term has degree one, and the equation would be homogeneous. With me? Here is the beautiful idea. Set each bracket equal to zero, and you get two straight lines. If small a times b prime is not equal to a prime times b, the lines are not parallel, so they meet at one point, h comma k. The equals sign with a slash through it means not equal to. Clear?

1:33Case 1: shift the origin to (h,k)

Now, case one, when the lines meet. We move the origin to h comma k: put x equals capital X plus h, and y equals capital Y plus k. These are the new coordinates, measured from that point. Okay? Let me ask you this: after this shift, does d y by d x change? It does not change at all, because h and k are constants. So d y by d x is simply d capital Y by d capital X, as simple as that. Put x equals capital X plus h, and y equals capital Y plus k, into the top bracket. The constant left over is small a times h, plus b times k, plus c, and that is zero, because h comma k lies on the first line! The same happens in the bottom. So d capital Y by d capital X equals small a times capital X, plus b times capital Y, over a prime times capital X, plus b prime times capital Y, which is homogeneous. See the magic? Then put capital Y equals v times capital X, as in the last lesson. And here is a common trap: the question was in small x and y. So at the end, put back capital X equals x minus h, and capital Y equals y minus k. Never stop one step early!

2:59Case 2: parallel lines

Now, case two. Suppose a prime over small a equals b prime over b. Then the two lines have the same slope: they are parallel. Can we still shift the origin? We cannot, because parallel lines never meet. But now the x and y part of the bottom is a multiple of small a times x plus b times y, so the same block appears twice, just like the substitution from last chapter. So put t equals small a times x, plus b times y. Then t prime equals small a, plus b times y prime. Never drop that small a! Now only t and x remain, and the equation is separable. Clear?

3:42Special case: b = -a' makes the equation exact

One special case to spot. Cross multiply, and the equation reads: the top bracket times d x, minus the bottom bracket times d y, equals zero. Okay? If b equals minus a prime, this equation is exact. You will study exact equations in Chapter five; such an equation integrates at once, with no substitution. Right?

4:06Example 1: (x + y + 1)/(2x + 2y + 3), put t = x + y

Now, Example one. Solve d y by d x equals x plus y plus one, over two x plus two y plus three. Which case is it? Pause the video and try it first. Done? Okay, let's check. Both ratios are two, so the lines are parallel, and we put t equals x plus y. Then t prime equals one, plus the fraction t plus one over two t plus three. Simplify, and that is three t plus four, over two t plus three. With me? Separate the variables. The tall S shape is the integral sign. The integral of two t plus three, over three t plus four, d t, equals the integral of d x, assuming three t plus four is not zero. So how do we integrate this fraction? Split it like a division: two thirds, plus one third over three t plus four. Now integrate: two thirds times t, plus one ninth times the natural log of mod three t plus four, equals x plus small c. The two straight bars are the modulus, and the extra third comes from the three in front of t. Okay? Finally, multiply by nine, and nine times small c becomes a new constant, called capital C, right? Put t back as x plus y: six x plus six y, minus nine x, leaves six y minus three x. So the answer is six y minus three x, plus the natural log of the modulus of three x plus three y plus four, equals capital C, and we are done!

5:50Example 2: (x + 2y - 3)/(2x + y - 3), shift origin to (1, 1)

Now, remember our equation from the start? Let's finish it. Example two: solve d y by d x equals x plus two y minus three, over two x plus y minus three. Pause the video and try it first. Done? Okay, let's check. The ratios are two and one half, so the lines are not parallel. So where do they meet? Solve x plus two y equals three, and two x plus y equals three, and you get the point one comma one. So put capital X equals x minus one, and capital Y equals y minus one. Clear? The constants vanish, and d capital Y by d capital X equals capital X plus two capital Y, over two capital X plus capital Y, which is homogeneous. Put capital Y equals v times capital X, and the left side becomes v plus capital X times v prime. Move v across: capital X times v prime equals one minus v squared, over two plus v. Got it? Separate, assuming one minus v squared is not zero, and use partial fractions. One minus v squared is one minus v, times one plus v. Put v equals one, and the first constant is three halves. Put v equals minus one, and the second constant is one half. Right? Integrate both sides. One half natural log of mod one plus v, minus three halves natural log of mod one minus v, equals natural log of mod capital X, plus small c. Watch that minus sign; it comes from the minus v inside. Clear? Multiply by two and remove the logs: one plus v, over one minus v whole cubed, equals capital C times capital X squared. Put v equals capital Y over capital X, cancel capital X squared, and capital X plus capital Y equals capital C times capital X minus capital Y, whole cubed. Nice, right? Now go back to small x and y. Capital X plus capital Y is x plus y minus two, and capital X minus capital Y is x minus y. So x plus y minus two equals capital C times x minus y, whole cubed. And those two minus threes from the start vanished, just by moving the origin!

8:19Quick revision: lines meet, lines parallel, b = -a' exact

So, quick revision. If the two lines meet, shift the origin to their meeting point, put capital Y equals v times capital X, and at the end go back to x and y. If the lines are parallel, put t equals small a times x plus b times y, and separate. And if b equals minus a prime, the equation is exact and integrates at once. Now go and try the practice problems on this page. All the best!