Solutions of a Differential Equation

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Three students solve the same equation: d y by d x equals two x. The first one writes y equals x squared. The second writes y equals x squared plus two. And the third writes y equals x squared minus seven. So who is right? Well, all three of them are right! Differentiate any one, and you get two x every time. So one equation has many answers. Then which answer does a question actually want? Today we sort this out: the general solution, the particular solution, and a strange third kind, the singular solution. These ideas come up often in JEE, so let's begin.

0:46What is a solution?

So, first question: what exactly is a solution? The letters D E on screen are short for differential equation. Now, a solution is a function that satisfies the equation identically. That means it works for every value of x, not just at one or two lucky points. Okay? Now look at the three answers from our puzzle, in the box: x squared, x squared plus two, and x squared minus seven. What do they have in common? Yes, exactly! They differ only by a constant, and the derivative of a constant is zero. The fraction d y by d x is the derivative of y with respect to x. So for d y by d x equals two x, y equals x squared plus capital C works for every constant, whatever its value. That gives infinitely many solutions from just one equation. With me?

1:43The general solution: a whole family

Now look at the picture. This is the family y equals capital C times x squared, from our last lesson. Each curve comes from one value of capital C, and every curve satisfies x times y prime equals two y, as the caption says. The small dash after y is called a prime, and y prime means d y by d x. Okay? Here is something to notice. This equation is first order, because only the first derivative appears. And the family has exactly one constant. Is that just a coincidence? No, it is not a coincidence; it is the rule! The general solution of an equation of order n contains n independent arbitrary constants. Independent means they cannot be merged into fewer constants, as we saw last lesson. First order, one constant; second order, two constants. Clear? So the whole family is the general solution. Now pick just one curve, say one of the orange ones. Fixing the constant like this gives a particular solution, one single curve. Got it? And how do we fix the constants? Usually from an initial condition: y of x naught equals y naught. The small zero written below x and y is a subscript, so x naught is just one fixed value of x. The condition says the curve must pass through the point x naught, y naught. An equation together with such a condition is called an initial-value problem. Right?

3:16How many conditions do we need?

Let me ask you this. Here is y double prime equals four y. The two dashes mean the second derivative, so this equation is of order two. How many initial conditions do we need to fix a particular solution? Yes, we need two of them. Order two means two arbitrary constants, and each condition fixes one of them. For example, the value y of zero, and the slope y prime of zero. So remember: one condition for each constant.

3:47Is every solution in the family?

Okay? Now a natural question. The general solution has all those constants. So surely every solution comes from it, just by choosing the constants, right? Not always, so be careful here! Some equations also have a singular solution. It satisfies the differential equation, okay? But no value of the constants gives it from the general solution. It lives outside the family. We will catch one in Example two. With me?

4:16Example 1: particular solution of y' = 2x with y(1) = 3

Now, Example one. Find the particular solution of y prime equals two x, with y of one equals three. That means when x is one, y must be three. Pause the video and try it first. Done? Okay, let's check. Integrate two x, and you get y equals x squared plus capital C, the constant of integration. First order, so exactly one constant, just as the rule says. This is the whole family from our puzzle. Okay? Now many students drop the constant here and write y equals x squared. Is that the answer? Does it pass the condition? No, it does not pass the condition, because at x equals one, x squared gives one, not three. So keep capital C, and use the condition: put x equals one and y equals three. One plus C equals three, so C equals two. Right? So the particular solution is y equals x squared plus two. And look, that is the second student from our puzzle! All three answers solved the equation, but only this one passes through the point where x is one and y is three. That is the job of an initial condition. Clear?

5:36Example 2: singular solution of y = xy' + y' squared

Now a harder one, Example two. Show that y equals c times x, plus c squared, is the general solution of y equals x times y prime, plus y prime squared. Then show that y equals minus x squared over four is also a solution. Is it a particular solution? Pause the video and try it first. Done? Okay, let's check. For y equals c times x plus c squared, y prime is just c, the constant. Put this into the right-hand side: x times c, plus c squared, which is exactly y. First order, one constant, so this is the general solution. Okay? Now the parabola, y equals minus x squared over four. Its derivative, y prime, is minus x over two. With me? Now put these into the right-hand side. Here, x times y prime gives minus x squared over two, and y prime squared gives plus x squared over four. Together they make minus x squared over four, which is y again. So the parabola is also a solution. Right? So it solves the equation. Then it must be a particular solution, correct? Tell me, which value of c gives this parabola? So let us look closely at the family. For each fixed c, c times x plus c squared is a straight line. But minus x squared over four is a parabola, and no straight line can ever be a parabola. Clear? So no value of c gives it. It is not a particular solution; it is a singular solution, the third kind we just met. And that is the trap: a solution need not be a member of the family. Got it? And here is one more beautiful fact. This parabola is the envelope of the lines y equals c times x plus c squared, which means it touches every one of those lines. So the final answer is no; it is a singular solution.

7:40Quick revision: general, particular and singular solutions

So, quick revision. The general solution of an equation of order n has n independent arbitrary constants. Fix them with initial conditions, like y of x naught equals y naught, and you get a particular solution. And a singular solution satisfies the equation, but no value of the constants gives it. Now go and try the practice problems on this page. You can do them. All the best!