Topic 7.2: Verifying Solutions for Differential Equations

The referee

That rule is so reliable that we can use it as a referee, to check whether a formula for somebody’s speed could really be true. Meet three simulated skydivers falling through our still air. Each one falls according to its own formula for its speed, and all three start from rest at the same instant.

Ghost A has no air at all, so it gains ten meters per second every second forever, which can’t happen in air. B and C look believable, since both start slow and level off right at fifty. To referee them, take a ghost’s speed at one instant, ask the rule how fast that speed should be changing, and compare it with how fast the ghost’s own formula changes it. At the start all three pass, ten and ten.

Ghost B starts out perfect, but I don’t trust it. Drag the clock and find a moment when B’s two arrows disagree by at least two meters per second squared. Ghost C is different. Its two arrows sit exactly on top of each other at every moment, and that’s what it means to obey the rule.

We call a formula like C’s a solution, and finding one is what solving a differential equation means. A function is a solution when you substitute it and its derivative, and the two sides come out equal for every t, not just one.

Let’s check ghost A on paper. Its formula is v equals ten t, so the left side, its derivative, is ten. On the right we put ten t in for v and get ten minus point four t squared. Ten equals ten minus point four t squared only when t is zero, so A passes at one instant and fails at every other, and v equals ten t is not a solution.

B is harder to check with algebra, so let’s test one time, five seconds. Its speed then is about thirty-one point six. By the chain rule its derivative is ten e to the negative t over five, and at five seconds that’s about three point six eight.

Now you be the referee. What does the rule demand at five seconds? Work out ten minus point zero zero four times thirty-one point six squared, to the nearest tenth.

One mismatch is enough to throw a formula out, but calling something a solution means the sides match for every t, and that takes algebra, not a few test points. Ghost C’s formula passes that algebra too, so C is the real solution here.

Newton’s iron

Rules like this one are a lot older than skydiving. In a paper he published in 1701, Isaac Newton described heating a block of iron until it glowed, taking it out of the fire, and letting it cool in a steady wind.

If you took precalc with us, you’ve stood in this yard. Newton’s rule says the iron loses heat faster the hotter it is compared with the air. The needle shows how fast the heat drops. The bracket shows how much hotter the iron is than the air. Halve one and the other halves.

Let’s leave Newton’s scale and write a rule like his the way the exam asks, in degrees Celsius. A hot bolt sits in a twenty-degree workshop, and its temperature T changes at a rate proportional to the difference between T and twenty. The rate is d T d t, proportional to means a constant k times, and the difference is T minus twenty.

The bolt is cooling, so the rate should be negative while T is above twenty, and with k positive that’s d T d t equals negative k times T minus twenty. For a cold bolt below twenty, T minus twenty is negative, so the rate turns positive and it warms up, as it should.

Two more phrases come up a lot. If a population P doubles and its growth rate doubles too, the rate is proportional to P, so d P d t equals k P. Inversely proportional means doubling y halves the rate, so y goes in the denominator, and d y d t equals k over y.

Water drains out of a tank so that the depth h falls at a rate proportional to the square root of the depth. For a positive constant k, which equation says that?

Now a full verification, the way you’d write it on the exam. The bolt’s rule is d T d t equals negative one fifth of T minus twenty, so here k is one fifth, with t in minutes, and someone claims T equals twenty plus eighty e to the negative t over five. First the left side. The derivative of the twenty is zero. For the exponential part, the chain rule brings in the inner derivative, negative one fifth.

Eighty times negative one fifth is negative sixteen, so the left side is negative sixteen e to the negative t over five. Now the right side. We put the whole formula in for T, the plus twenty and the minus twenty cancel, and negative one fifth of eighty is negative sixteen, so the right side is negative sixteen e to the negative t over five too.

The two sides are the very same expression, so they match at every time, and that’s a verified solution. It makes sense, too, since the bolt starts out dropping sixteen degrees a minute and slows down as it nears room temperature.

On the exam, finish with a sentence. Because both sides equal negative sixteen e to the negative t over five for all t, T satisfies the differential equation. If they give a starting temperature of a hundred, check it too, and T of zero is twenty plus eighty, a hundred.

One rule, many futures

Let’s check a second formula together. Q equals a hundred e to the negative t over five also starts at exactly a hundred degrees. Its derivative is a hundred times negative one fifth, so negative twenty e to the negative t over five. On the right, negative one fifth multiplies both pieces.

The exponential part becomes negative twenty e to the negative t over five, and the minus twenty becomes plus four. The sides are off by four degrees a minute at every time, so Q is not a solution, even though it starts at the right temperature. It’s a cooling model with no floor, one that would cool the bolt right past room temperature.

Back in Newton’s yard, heat is on his own scale, where the air is four and boiling water is about thirty-four. Let’s set two more blocks beside the iron, one out of a kettle of boiling water and one out of a bucket of melting snow, which is colder than the air. Under the same rule, the iron and the kettle block cool while the snow block warms, and all three curves bend toward the temperature of the air, every needle matching the rule.

Now start a block of your own. Drag its starting heat and find the one block whose temperature never changes at all. Now you tune the formula. Yours is four plus C times two to the minus r t, and r is how fast it cools, so turn r until every needle agrees with Newton’s rule.

Every start works, but only one cooling speed does. The rule fixes how fast the heat falls, and the start picks which curve is yours. Back to the bolt, let’s check a whole family at once by putting a letter where the eighty was, T equals twenty plus C e to the negative t over five. The left side is negative C over five e to the negative t over five, and on the right the plus twenty and minus twenty cancel and leave exactly the same thing.

That works for every C, so one differential equation has infinitely many solutions. C equals eighty is our hot bolt, C equals zero is a bolt that stays at twenty forever, like the skydiver stuck at fifty, and a negative C is a cold bolt warming up. The equation is the rule that every one of these follows, and the starting temperature picks out which curve is yours.

On the exam

Try this one on your own. Which of these functions is a solution of d y d x equals y minus x? The same check works when an equation involves the second derivative. For y double prime plus nine y equals zero, try y equals cosine of three x. Its second derivative is negative nine cosine of three x, so adding nine y gives zero for every x.

Before we leave the board, look at where students lose points. Checking a formula at one point, or only at its starting value like Q, doesn’t verify it. If you drop the chain rule’s inner factor, a real solution will look wrong, so take that derivative carefully.

Pull the chute

Let’s finish where we started. Just before he opens his parachute, Felix is doing a hundred eighty-five kilometers an hour. That’s right at the top speed for this thick air, so the right side of his equation is almost exactly zero. When the canopy opens, k jumps, so the top speed, the square root of g over k, drops. Real parachutes are made so you land at about eighteen kilometers an hour, which is five meters per second.

So now you design his canopy. Set k so the equation’s top speed comes out at five meters per second, then let go and watch him land.

Here is his real canopy. His speed drops from a hundred eighty-five to about eighteen kilometers an hour in a few seconds. A k like yours is about a hundred times his free-fall k, so the top speed shrinks by the square root of a hundred, a factor of ten. He floats the rest of the way down.

That whole jump ran on one rule, d v d t equals g minus k v squared, with k growing as the air thickened. You used its zero to explain his record speed and to design his canopy, without ever solving it.

You caught a formula that looked right by finding where its derivative broke the rule, and you wrote your own cooling formula and found that one rule has infinitely many solutions. That’s what a differential equation is, a rule for how something changes. In the next lessons you’ll learn to sketch its solutions and then to solve for them.