Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Two students solve the same linear differential equation, each starting from a different point, and draw both solution curves on one graph. Can the two curves cross each other? Most people say yes; curves cross all the time. Hold that thought; you will prove the answer yourself soon. Today is Part C of our problem bank: quick concept checks, the kind of ideas JEE can test in statement form, so let's begin.

0:32Why exactly one arbitrary constant?

So, our first concept check. The general solution of a first order equation contains exactly one arbitrary constant. Why is that? First, the rule from chapter one. The order of an equation is its highest derivative. And the general solution of an equation of order n contains n arbitrary constants. Okay? Now the reason. First order means only d y by d x appears, so solving it amounts to one integration. And each integration brings in one arbitrary constant, the plus capital C, right? And one condition, such as y of x naught equals y naught, fixes that constant. The small zero below x and y is a subscript, so x naught is one fixed value of x. One condition picks out one single curve. Clear?

1:26Circles of fixed radius: what order?

Now the next check: the order of the differential equation of all circles of radius r, with r fixed. Each circle is x minus h, the whole squared, plus y minus k, the whole squared, equals r squared. Here h and k fix the centre, and r is the radius. With me? Now recall: if a family has n essential constants, it gives a differential equation of order n. The double arrow on screen means, gives. Essential means they cannot be merged into fewer. Okay? So let us count. I can see three letters here: h, k and r. So three constants, and the order is three, right? No, no, be careful here, because the radius is fixed, the same for every circle. Only the centre, the point h comma k in round brackets, can move, so there are just two constants. Got it? So the order is two. You would differentiate twice and eliminate h and k. The trap was counting letters instead of reading what is fixed. Right?

2:32Linear, homogeneous, or both?

Now a classification check. Here is y prime equals x squared plus y squared, over x times y. The small dash after y is called a prime, and y prime means d y by d x. So, is it linear, homogeneous, or both? Let us test it. The linear form from chapter four is y prime plus P of x times y, equals Q of x, where y appears only to the power one. Okay? Now homogeneous: the top and the bottom must be homogeneous of the same degree. The top, x squared plus y squared, and the bottom, x times y, both have degree two. So yes, it is homogeneous. With me? But is it linear? No, because it contains y squared on top, and the y in the bottom gives one over y. Neither fits the linear form. Clear? Now here is a bonus for you. Split the fraction: y prime equals x over y, plus y over x. Move y over x to the left: y prime minus one over x times y, equals x times y to the power minus one. That is Bernoulli's form, with n equal to minus one, which is neither zero nor one. So it is a Bernoulli equation too. Got it?

3:51Example 1: is y = x squared the general solution of xy' = 2y

Now Example one. Is y equals x squared a solution of x times y prime equals two y? And is it the general solution? Pause the video and try it first. Done? Okay, let's check. If y equals x squared, then y prime equals two x. Okay? Now the left side: x times y prime is x times two x, which is two x squared. The dot on screen means times. And the right side, two y, is also two x squared. They match for every x, so yes, it is a solution. Clear? It satisfies the equation, so it must be the general solution, right? Oh no, careful here, because this equation is first order, and from our first check, its general solution must contain exactly one arbitrary constant. But y equals x squared has no constant at all. Got it? The general solution is y equals capital C times x squared. And it works, since x times y prime is two times capital C times x squared, which is two y. The picture shows this family, one curve for each capital C value. With me? So y equals x squared is just the particular solution with capital C equal to one. The answer is yes, and no. Notice also that every curve in the picture passes through the origin. Remember that for Example two, okay?

5:22Example 2: can two solutions of a linear DE ever cross?

Now Example two: our puzzle from the start. Take the linear equation y prime plus P of x times y, equals Q of x, where P and Q are continuous. Can two different solutions cross? Pause the video and try it first. Done? Here is a hint. If two solutions meet at some point, what is their difference there? Yes, exactly, the difference is zero! So call the two solutions y one and y two; the small one and two below y are subscripts. Let u be y one minus y two. Subtracting gives u prime plus P times u equals Q minus Q, which is zero. With me? Now multiply by the integrating factor, e to the power integral P d x; the long S is the integral sign. Then the derivative of u times the integrating factor is zero, so u times the integrating factor is a constant, capital C, and u equals capital C times e to the power minus integral P d x. Okay? Now the key point. The dots in the bracket stand for any power: e to any power is always greater than zero, never zero. So u can be zero at some point only if capital C is zero. And then u is zero everywhere: y one and y two are the same solution. Right? So the answer to our puzzle is no: two different solutions never cross, or even touch. The picture shows three solutions of y prime equals x minus y, namely y equals x minus one, plus capital C times e to the power minus x, for capital C equal to two, zero and minus one; the short grey strokes show the slope at each point. The curves come close, but never meet. Got it? But wait a moment: in Example one, every curve of y equals capital C times x squared passed through the origin. And x times y prime equals two y is a linear equation too. So did our rule just break? No, it did not, and here is why. Divide by x to get the standard form: y prime minus two over x times y, equals zero. So P is minus two over x, which is not even defined at x equals zero. Our result needs P and Q to be continuous, and at the origin that fails. That is why we always read the conditions. Clear?

7:55Quick revision: constants, order, type, no crossing

So, quick revision. Order n means n arbitrary constants; first order, one integration, one constant. For circles of fixed radius, only the centre is arbitrary, so the order is two. Our fraction equation is homogeneous and Bernoulli, but not linear. And y equals x squared is only the member with capital C equal to one, not the general solution. For a linear equation with P and Q continuous, two different solutions never cross. Now try the two practice problems on this page. All the best!