The mirror-image form, where x is treated as the dependent variable

0:00Every differential equation you have solved so far has had y as the unknown and x as the letter it was differentiated against. That looks like a fact about the alphabet. It is not. It is a choice, and nobody ever told you it was being made. An equation relates two quantities. Deciding which one is the unknown is a decision you make when you look at it, and you can make it either way.

0:29This video is about making it the other way, on purpose, and about the two places where doing that is the difference between an equation you can solve in six lines and one you cannot start at all. And there is a second reason to care, which is harder to see. Some of these equations arrive already written the other way round, and are not the ones that need it.

0:57The move itself is one line. Turn the derivative over. The derivative of y with respect to x, and the derivative of x with respect to y, are reciprocals of each other. That is a fact about the curve, not about the notation, so it is worth checking rather than believing. Three relations were taken here, each written twice: once as a rule giving y from x, and once as a rule giving x from y. Neither rule was obtained by inverting the other symbolically. Each was written down on its own.

1:36Then both slopes were measured, from difference quotients on both sides of the point, and multiplied together. At all fifteen places the product is one. Aim the same reading at a target a fiftieth away and it lands at none of them. There is one place where the turn is not available, and it is worth knowing where. If the forward slope is nought, its reciprocal does not exist. Take a cube, and read its slope at the origin. Measured from both sides, it settles on nought exactly.

2:13So at that one point the curve has a horizontal tangent, the mirrored slope would have to be infinite, and there is nothing to turn over. Everywhere else on that curve, the swap is legal. That is the condition, and it is the only one. Now the first of the two methods, and it comes with a gift. You have already met expressions that are homogeneous: scale both variables by the same factor and the whole expression picks up a fixed power of that factor.

2:47What is easy to miss is that every one of them can be written two ways. A power of x, times a function of y over x. Or the same power of y, times a function of x over y. Take y squared plus twice x y. Pull out x squared and what is left is a function of the ratio y over x. Pull out y squared instead and what is left is a function of the ratio the other way up.

3:19Both were measured at eight places. Both hold at all eight, and neither meets a target a hundredth away anywhere. And the expression that is homogeneous at no degree factorises neither way. The sine of x added to the cosine of y. Scale both variables and nothing comes out in front. Put it through the same factorisation test and it lands at nought places for a power of x, and nought places for a power of y.

3:52That matters, because a test that only ever agrees is not a test. The one that fails is what makes the three that pass mean something. So the symmetry is real, and it is the reason the substitution has a mirror image at all. Here is the mirrored substitution. Forward, you set y equal to a new letter times x. Mirrored, you set x equal to a new letter times y.

4:22The new letter multiplies the other variable. That is the whole difference, and it is the thing most often got backwards. And now something that is worth saying carefully, because it corrects a natural assumption. For an equation whose right hand side is homogeneous of degree nought, both substitutions separate it. Degree nought means the expression takes one value all along every ray from the origin, and that is true whichever letter you make the multiple.

4:57So the mirror is a choice about which is less work. It is not a choice about which is legal. And you can see that the degree is what buys it, by taking it away. A degree-one right hand side was put through the same test, both ways round. With the new letter held fixed, what is left after the substitution still carries the other variable. Both ways. Neither separates.

5:25So the condition is doing work, and it is doing the same work in both orientations. Which brings us to the standard worked example, and it is the cleanest mirrored item in the whole topic. Twice y times an exponential of x over y, against dx, plus a bracket holding y minus twice x times that same exponential, against dy, equal to nought. And a data point: x is nought when y is one.

5:59Read it with x as the unknown. The derivative of x with respect to y is twice x, times the exponential, minus y, all over twice y times the exponential. Measured by scaling, that is homogeneous of degree nought. So it qualifies. Substitute x equal to a new letter times y. The product rule gives the new letter plus y times its derivative. Everything collapses. What is left is twice an exponential of the new letter, against the reciprocal of y, with a minus sign.

6:41Integrate both sides. Twice the exponential of the new letter, plus the logarithm of the modulus of y, equals a constant. Put the new letter back. Twice the exponential of x over y, plus the logarithm of the modulus of y, equals a constant. Now the data point. At x nought and y one, the exponential of nought is one and the logarithm of one is nought, so the left side is exactly two.

7:14The constant is two, and the particular solution is that relation. And it was checked the only way worth checking: the relation was solved for x on a branch, turned back into a curve, and the slope of x against y measured from difference quotients at five places. It closes the equation at all five, and closes a target a fiftieth away at none. One thing to notice about that answer, because it is a question people ask.

7:48There is nothing to convert back. A relation between x and y is a relation between x and y. Which letter you called the unknown was bookkeeping while you were solving; it leaves no trace in the answer. Every mirrored example stops the moment the relation is found, and that is not laziness. There is genuinely nothing left to do. Now the second method, and here the mirror earns its keep properly.

8:21The linear shape, forward, is the derivative of y plus a function of x times y, equal to another function of x. Mirrored, it is the derivative of x with respect to y, plus a function of y times x, equal to another function of y. Every symbol swaps. Both coefficients are now functions of the other letter alone. And the recipe swaps with them. The factor is the exponential of the integral of the first coefficient, integrated against y. Then the unknown times the factor equals the integral of the second coefficient times the factor, with respect to y, plus a constant.

9:08Two things in that sentence are worth stopping on. The first is the integral sign. It is easy to write that exponent without one, and a version without it circulates. The factor is an exponential of an integral, not an exponential of a product. The second is the letter you integrate against, and this is where the marks go. The factor is built by integrating the first coefficient against y. Not against x.

9:41Get that wrong and you produce a factor that is a function of the wrong variable, and nothing after it works. This was measured. The right factor satisfies the condition the derivation demands at all five places. A factor built by integrating against the wrong letter satisfies it at none. The shortest complete item in the topic runs six lines. y times dx, minus a bracket holding x plus twice y squared, times dy, equal to nought.

10:18Divide by y and by dy. The derivative of x with respect to y, minus x over y, equals twice y. So the first coefficient is minus the reciprocal of y. Its integral against y is minus the logarithm of y, and the factor is the exponential of that, which simplifies to the reciprocal of y. Step three. x over y equals the integral of twice y times the reciprocal of y, which is the integral of two, which is twice y, plus a constant.

10:57So x equals twice y squared, plus a constant times y. Measured at five places, it closes the equation at all of them. And that example is where the mirror stops being a convenience. Look at that equation read the forward way. The derivative of y with respect to x equals y over a bracket holding x plus twice y squared. The unknown is now inside a denominator, and it is squared in there.

11:31Put both readings through the linearity test and the verdict is clean. Read for x, the equation is linear: the test holds at all twenty readings. Read for y, it fails at the first reading it takes. Which means that read the forward way there is no integrating factor to find, because the method does not apply. The mirror was not a shortcut. It was the only route. The hardest item in this topic combines the mirrored recipe with two integration techniques, and it is worth watching once.

12:10A bracket holding an inverse tangent of y, minus x, times dy, equal to a bracket holding one plus y squared, times dx. Rearranged, the derivative of x with respect to y, plus x over one plus y squared, equals the inverse tangent of y over one plus y squared. The first coefficient is the reciprocal of one plus y squared. Its integral against y is the inverse tangent of y, so the factor is the exponential of the inverse tangent.

12:50On the right you get the inverse tangent, over one plus y squared, times that exponential. Set a new letter equal to the inverse tangent and the whole thing becomes that letter times its own exponential, which yields to parts. The answer is x equals the inverse tangent of y, minus one, plus a constant times the exponential of minus the inverse tangent. Measured at five places, it closes at all five.

13:25So how do you decide, in front of an equation you have not seen before? Not by how it is written down. By which letter it is linear in. Linear in a letter means that letter and its derivative each appear to the first power, and neither of them appears inside anything else. That is testable, and it was tested here by superposition rather than by looking. Read the equation both ways. Write each reading as an operator, feed it a combination of two curves, and ask whether what comes out is the combination of what the two curves gave separately.

14:06The orientation that passes is the orientation to take. And notice how the two verdicts differ in kind: crediting the linearity took twenty readings, because it is a claim about every place. Refusing it settled on the first, because one disagreement is enough. Now the two traps, and they point in opposite directions. Three of the general-solution questions you will meet are set the forward way, with the derivative of y on top, and every one of them needs mirroring. Each is linear in x and not linear in y. A student who does not know to invert cannot start any of them.

14:52And one of the closing questions is set the other way, with the derivative of x on top, and is not a mirrored problem at all. Read for y it is an ordinary linear equation. Read for x, the unknown sits inside a square root and inside an exponential, in the coefficient position that has to hold a function of the other letter alone. The test refuses it in that orientation immediately.

15:23So: the layout is not the structure. Read what the equation is linear in, and ignore how it was set. One more thing, and it is the sharpest point in this topic. The mirrored substitution is defined for equations whose right hand side is homogeneous of degree nought. But the substitution is wider than the class it was defined for. Here is an instance built to show it. The derivative of x with respect to y equals x over y, plus x squared over y, plus y.

16:02Scale both letters and the three terms pick up three different powers of the factor. Measured, that expression is homogeneous at no degree at all. It is outside the class the substitution was defined for. Substitute x equal to a new letter times y anyway. With the new letter held fixed, the leftover still carries y, so the condition genuinely fails. And it separates regardless, because the leftover comes apart into y times one plus the new letter squared. Measured at all twenty combinations, and nowhere against a target a hundredth away.

16:47The answer is x equals y times the tangent of y plus a constant, and it closes that equation at all five places tested. So the moral is precise. If the substitution works, that does not prove the equation was homogeneous. Reasoning backwards from a tool to the class it was defined for is a mistake, and it is one nothing prepares you for, because a method is usually introduced as though its definition and its reach were the same thing.

17:22They are not. Use the test for homogeneity to decide whether the method is guaranteed. Do not use the method working as evidence that the test would have passed. A practical warning before the end. A revision summary will carry the mirrored form for the homogeneous method. It gives both derivatives and applies the degree condition to each. For the linear method it does not. It gives the forward shape only, states the condition on the two coefficients, and stops.

17:59Meanwhile five questions across two exercise sets need exactly the half that was dropped. Four of those five are solved and measured in the working behind this video. So if you revise from a summary and meet a linear equation that needs mirroring, nothing on that page will have told you the mirror exists. That is not a reason to distrust summaries. It is a reason to know which half of this topic is yours to remember.

18:33What to carry away. Which letter is the unknown is a choice. The two derivatives are reciprocals wherever the slope is not nought, and that is the whole licence for making the choice either way. For a homogeneous equation, either substitution separates it, so pick the one that leaves less work. For a linear equation, only one orientation has a factor at all, so pick the letter the equation is linear in.

19:05Decide that by testing, not by reading the layout. Three set the forward way need mirroring; one set the mirrored way does not. The mirrored factor is built by integrating against the other letter. Everything else in the recipe is the recipe you already know. And when you find an answer, stop. A relation between x and y does not remember which letter you called the unknown.