Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle: d y by d x equals three x squared times e to the power minus y, and y is zero when x is zero. Find y when x is two. Two students solve it. One writes the natural log of eight, the other, the natural log of nine. They cannot both be right, can they? So who slipped? This is Set one of our mixed problems on differential equations, and this style comes up often in JEE Main. So, let's begin.

0:33Set 1: the usual pattern

So, here is what Set one holds: eight solved problems in the JEE Main style, mixed from the whole unit. Order and degree, forming an equation, separable, linear, and decay. Okay? And most of them follow one pattern, the course says: an initial-value problem. Step one: name the type, whether it is separable, homogeneous, or linear. Right? Now, before step two, a question for you. The equations in this pattern are first order, because the highest derivative is d y by d x. So how many arbitrary constants does the general solution carry? That is correct, just one. So step two is to solve it. An n-th order equation has n arbitrary constants, so first order means exactly one, capital C, and not one on each side, okay? Step three: fix C at once with the given condition. Here, y of zero equals zero, in round brackets, means y is zero when x is zero. Clear? And step four: evaluate. It may be a value of y, an area, a limit, or an extremum, meaning a largest or smallest value. Today, both problems ask for a value. Got it?

1:48Tool 1: separate the variables

Tool one, from Chapter two: separating the variables. When d y by d x equals g of x times h of y, a part with only x times a part with only y, we can separate them. Okay? Divide by h of y: d y over h of y on the left, and g of x, d x, on the right. Integrate both sides; the long S is the integral sign. Plus capital C goes on one side only. Right? But wait. To bring h of y under d y, we divided by it. So what did we quietly assume there? Yes, exactly, we assumed that h of y is not zero. The equals sign with a slash through it means not equal to. Clear? So if h of y naught is zero, where y naught is the y with a small zero below it, the constant solution y equals y naught can be lost. Check it separately.

2:46Tool 2: the integrating factor

With me? Tool two, from Chapter four: the linear equation. d y by d x plus capital P times y equals capital Q, with P and Q functions of x only. The coefficient of y prime must be one; y prime, the y with a small dash, is d y by d x. Okay? Then the integrating factor, shortened on screen to its initials, is e to the power integral P d x. And here, we add no constant at all. Right? Multiply through by it, and the left side becomes the derivative of y times the integrating factor. So y times the integrating factor equals the integral of Q times the integrating factor, d x, plus capital C, okay? The dot means times. See where C sits: before we divide by anything. Clear? Now a quick one for you. If P is two over x, what is the integrating factor? Yes, x squared! The integral of two over x is two log x, and e to the power two log x is x squared. So here is a shortcut: P equals n over x gives x to the power n; the double arrow means gives. Got it?

3:55Example 1: separable, dy/dx = 3x squared e^-y, find y(2)

Now the first problem, an easy one: problem number two from the set, and yes, it is our puzzle. The equation is y prime equals three x squared times e to the power minus y, and y of zero is zero. Find y of two. Pause the video and try it first. Done? Okay, let's check. Three x squared is a function of x only, and e to the power minus y is a function of y only. So which tool do we use? Tool one, so we separate the variables! Dividing by e to the power minus y is safe, since it is never zero, so no solution is lost. That gives e to the power y, d y, equals three x squared, d x. Okay? Integrate both sides: e to the power y gives e to the power y, and three x squared gives x cubed. So, e to the power y equals x cubed, plus capital C, with me? Now, here is where one student slipped: no constant, just e to the power y equals x cubed. Then at x equals two, the answer is the natural log of eight. Looks fine, right? No, no, careful, check the condition. Put y of zero equals zero, and e to the power zero equals zero plus C, okay? But e to the power zero is one, so capital C is one, not zero. So the particular solution is e to the power y equals x cubed plus one. Clear? Now put x equals two. Two cubed is eight, and eight plus one gives nine. So y is the natural log of nine; l n on screen means the natural log. Puzzle solved: log nine is right, and the extra one came from the constant. If you got log nine, very well done!

5:45Example 2: linear, dy/dx = y/x + x, y(1) = 2, find y(2)

Now a harder one, problem number six from the set. The equation is y prime equals y over x, plus x, and y of one is two. Find y of two. Pause the video and try it, and here is a hint: use tool two. Done? Okay, let's check. First, the standard form. Bring y over x to the left side. Then what are capital P and capital Q? We get y prime, minus one over x times y, equals x. So P is minus one over x, and Q is x. Do not lose that minus sign. Okay? Now the integrating factor: e to the power minus the integral of d x over x, which is e to the power minus log x. Near x equals one, x is positive, the greater-than sign in brackets, so this is just one over x. Right? So y over x equals the integral of x times one over x, d x. That is the integral of one, d x, so y over x equals x plus capital C, okay? Now multiply by x to get y. Some students write y equals x squared plus C here. Then y of one equals two gives C equals one, and y of two comes out as five. Is five correct? No, no, wait, that is the trap! C is already on the right side, so it gets multiplied by x too. So y equals x times, x plus C, which is x squared plus C times x, and not x squared plus C, okay? Now use y of one equals two. One plus C equals two, so C equals one. So y equals x squared plus x. Got it? Finally, put x equals two. Four plus two, equals six. So the answer is six, not five: same constant, wrong place, different answer. Well done if you got six!

7:45Quick revision: name the type, fix C, add C before dividing

So, quick revision. Number one: name the type first, separable, homogeneous, or linear. Number two: first order means one constant, so fix it at once with the given condition. And number three: in a linear equation, the constant goes in before you divide by the integrating factor. Now go and try the practice problems on this page. All the best!