Equations Reducible to Separable Form

Transcript

0:00Introduction

Hello everyone! Let us start with a small challenge. Here is an equation: d y by d x equals x plus y, whole squared. In the last lesson we learnt to separate the variables, with all the y terms on one side and all the x terms on the other. So try it here. Can you separate x and y? Expand it, and you get x squared, plus two x y, plus y squared. That middle term mixes x and y together, so there is no way to split them. Stuck, right? Today one clever substitution will unlock this equation completely. This idea comes up often in JEE, so let's begin.

0:42When x and y come as one package

So, look at the pattern on screen. The small dash after y is a prime, and y prime means d y by d x, okay? Here y prime equals f of small a times x, plus b times y, plus c, where small a, b and c are constants. Now notice this: x and y never appear alone. They always come together as one package. With me? So let me ask you this. When the same package of x and y keeps coming back, what is the smart thing to do? Yes, exactly! Give the whole package one name. Put t equals small a times x, plus b times y, plus c. Now the right-hand side is simply f of t. Nice, isn't it? Now we differentiate t with respect to x. Many students write d t by d x equals f of t, straight away. Is that right? No, that is the trap, because t is not y. The derivative of small a times x is small a, and the derivative of b times y is b times y prime. So d t by d x equals small a, plus b times y prime, and y prime is f of t. That gives small a, plus b times f of t. Never drop that first term. Clear? Now look at this equation carefully. Is there any y left in it? Only t and x. So what can we do now? Yes, we separate the variables, just like last lesson! The integral of d t, over small a plus b times f of t, equals x plus capital C, the constant of integration. The tall S shape is the integral sign. And in the brackets, the equals sign with a slash through it means not equal to. We divided by small a plus b times f of t, so we assume it is not zero. Got it?

2:35Spot the package

Now let us train our eyes. Here is y prime equals sine of x plus y. What would you put t equal to? Yes, t equals x plus y. That is the package, with small a and b both equal to one, and c equal to zero. Easy, right? Now a fraction: y prime equals x plus y plus one, over x plus y minus one. Here x and y appear twice. So what is t now? That is correct, it is again x plus y. The top and the bottom use the same package, so one letter covers both. That is the whole skill: look for the same combination of x and y every time. Okay?

3:23Example 1: solve dy/dx = (x + y) squared with t = x + y

Now, remember our challenge from the start? Let's finish it. Example one: solve y prime equals x plus y, whole squared. Pause the video and try it first. Done? Okay, let's check. The package is x plus y, so put t equals x plus y. Then t prime equals one plus y prime. The one comes from the derivative of x, so do not drop it. And y prime is t squared, so t prime equals one plus t squared. Got it? Now separate the variables. The integral of d t, over one plus t squared, equals the integral of d x. One plus t squared is never zero, so dividing by it is completely safe. Now tell me, which standard integral is on the left? Yes, it is tan inverse t! The small minus one written up on tan means the inverse, not one over tan, right? So tan inverse t equals x plus capital C, okay? So, are we done here? Not yet, because the question is about x and y, not t. Put t back as x plus y, and take tan of both sides. So x plus y equals tan of x plus capital C, and that is the answer. The equation that looked stuck at the start is solved in a few lines! Clear?

4:50Example 2: dy/dx = root of (y - x) by two substitutions

Now a harder one, Example two. Solve y prime equals the square root of y minus x, for y greater than x. The tick shape over y minus x is the square root sign, and the sign in the brackets means greater than. Pause the video and try it first. Done? Okay, let's check. The package here is y minus x, so put t equals y minus x. Careful with the sign: t prime equals y prime minus one, and y prime is root t. So t prime equals root t, minus one. Right? We could separate this already, but the root makes the integral ugly. So how can we get rid of the root? Yes, with a second substitution! Put t equals s squared, where s is the positive square root of t. Then t prime equals two s times s prime, and root t is just s. So two s times s prime equals s minus one. Nice and clean, okay? Now we separate the variables. The integral of two s, d s, over s minus one, equals the integral of d x. Here we divided by s minus one, so we assume s is not equal to one. Keep that in mind; we come back to it at the end. With me? Next, how do we integrate two s over s minus one? We split the fraction first. Two s over s minus one equals two, plus two over s minus one. Now integrate each part, and the double arrow on screen leads to the result: two s, plus two times the natural log of mod s minus one, equals x plus capital C, the constant. The two straight bars around s minus one are the modulus. Got it? Now put s back as the root of y minus x. Two root of y minus x, plus two times the natural log of the modulus of root of y minus x, minus one, equals x plus capital C, and that is the general solution. Clear? And one last check, from the last lesson. We assumed s is not one, so check s equals one on its own. That means y minus x equals one, so y equals x plus one. Its y prime is one, and the root of y minus x is the root of one, which is also one. So y equals x plus one also satisfies the equation. Always check the zeros of what you divide by. Okay?

7:20Quick revision: put t = ax + by + c, keep the a in dt/dx

So, quick revision. When x and y always come as the same package, small a times x, plus b times y, plus c, put t equal to that package. Then t prime equals small a, plus b times f of t, and do not drop the small a. Separate, check what you divide by, integrate, and put t back in terms of x and y. Now go and try the practice problems on this page. You will enjoy them. All the best!