The Substitution y=vx

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle: x times d y by d x equals x plus y. Last chapter, we separated the variables, right? So try it here: can you get every y on one side, and every x on the other? You cannot, because x plus y does not split into a function of x times a function of y. So are we stuck? Not at all, because one clever substitution, y equals v x, makes it separable. Equations like this come up often in JEE, so let's begin.

0:38Homogeneous equations

So, look at this form. d y by d x, the derivative of y with respect to x, equals f of x, y, over g of x, y. Today's substitution works when f and g are of a special type, called homogeneous. With me? Here is what that means. Replace x by lambda x, and y by lambda y; lambda, the Greek letter on screen, is the number we scale by. If f of lambda x, lambda y equals lambda to the power n, times f of x, y, then f is homogeneous of degree n; the small n written up high is that power. And g must do the same. Clear? Now suppose f and g have the same degree, n. We scale x and y together by lambda. Tell me, what happens to the fraction, f over g? Yes, exactly, nothing happens to it! Lambda to the power n comes out on top and on the bottom, and cancels. So the right side depends only on the ratio, y over x. The double arrow on screen means this follows: d y by d x equals capital F of y over x. Such an equation is called homogeneous. Got it?

1:58The method: put y=vx

Now the method. Everything depends on y over x, so let us call that ratio v. So y equals v times x, where v is a new function of x. Now tell me, what is d y by d x? It is the product rule, because v also changes with x. d y by d x equals v, plus x times d v by d x. Oh no, be careful here, because many students forget the second term. Right? Put this into the equation. The left side becomes v plus x times d v by d x, and the right side becomes capital F of v. Move v across: x times d v by d x equals capital F of v, minus v. Only v and x are left. Okay? So the variables separate, just like last chapter. The tall curly S is the integral sign. The integral of d v over the bracket, capital F of v minus v, equals the natural log of mod x, plus capital C, the arbitrary constant. The letters l n mean the natural log, and the two vertical bars mean the modulus. Clear? And finally, never leave the answer in terms of v. Replace v by y over x, and you have y back. Got it? One more case. If the equation is homogeneous in the form d x by d y equals capital G of x over y, then put x equals v times y instead. Same method, with x and y swapped. With me?

3:34Trap: check homogeneity first

Now the trap. Look at d y by d x equals x plus y plus one, over x minus y. Is it homogeneous? Can we put y equals v x straight away? No, no, wait, we must check before we substitute. Scale x and y by lambda. The top becomes lambda x plus lambda y plus one, and that constant one gets no lambda at all. Right? So this equation is not homogeneous; the constant spoils it. Such equations need a shift of origin first, which comes in the next section. So always check homogeneity before you put y equals v x. Clear?

4:15Example 1: solve x dy/dx = x + y by putting y = vx

Now, Example one, and it is our puzzle from the start. Solve x times d y by d x equals x plus y. Pause the video and try it first. Done? Okay, let's check. Divide both sides by x, and d y by d x equals x plus y, all over x, which is one plus y over x. It depends only on y over x, so the equation is homogeneous. Right? Now put y equals v x. The left side becomes v plus x times d v by d x, and the right side becomes one plus v. Clear? Now look at both sides carefully. What happens to v? Yes, exactly, v cancels from both sides! We are left with x times d v by d x equals one, so d v equals d x over x. Okay? Integrate both sides. v equals the natural log of modulus of x, plus capital C, the constant. With me? Now replace v by y over x, and multiply by x. So y equals x times log of modulus of x, plus capital C times x. See, the equation we could not separate is solved in a few lines!

5:32Example 2: solve 2xy y' = x squared + y squared, y(1) = 2

Now Example two, a little harder. Solve two x y times y prime equals x squared plus y squared, given that y of one equals two, which means y is two when x is one. The small dash after y is called a prime, and y prime means d y by d x. Pause the video and try it first. Done? Okay, let's check. First, the homogeneity check. y prime equals x squared plus y squared, over two x y. Scale x and y by lambda, and the top and the bottom both pick up lambda squared. Both have degree two, so y equals v x will work. Right? Put y equals v x. On the right, x squared cancels, and v plus x times v prime equals one plus v squared, over two v. Move v across: one plus v squared, minus two v squared, gives one minus v squared. So x times v prime equals one minus v squared, over two v. Got it? Next we separate and integrate. Tell me, what is the integral of two v over one minus v squared? Is it simply the natural log of one minus v squared? Oh no, careful, there is a minus sign hiding here! The derivative of one minus v squared is minus two v, not two v. So when we separate, the integrals on screen give this: minus log of the modulus of one minus v squared equals log of modulus of x, plus small c. Clear? Bring the logs together and remove them. x times the bracket, one minus v squared, equals capital C, a new constant. Put v equals y over x, and multiply by x. So x squared minus y squared equals capital C times x, the general solution, with one arbitrary constant. Okay? Now use the condition. Put x equals one and y equals two. What is capital C? One minus four equals capital C times one, so capital C equals minus three. Did you get that? Put capital C equals minus three, and rearrange. y squared equals x squared plus three x, and that is our answer. Well done if you got it!

7:58Quick revision: same degree, y = vx, check homogeneity first

So, quick revision. An equation is homogeneous when d y by d x equals capital F of y over x, with top and bottom of the same degree. Put y equals v x, so d y by d x equals v plus x times d v by d x; then separate, and put v back as y over x. And a constant term spoils homogeneity, so check first. Now go and try the practice problems on this page. All the best!