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Separating variables and integrating, the constant an initial condition determines, the exponential model and its rate constant, and reading a slope field.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can solve a separable differential equation by separating the variables, integrating both sides and using an initial condition to pick out one solution from the family. You can recognise the equation whose rate is proportional to the amount, write down its exponential solution, find its rate constant from a doubling time or a half-life, and read the behaviour of a family of solutions off a slope field before solving anything.
You can antidifferentiate, and you have seen that $\int f\,dx$ names a family of functions rather than one. A differential equation is a question posed in that language: not 'what is this function's derivative?' but 'which function has this derivative?'
Differential equation: an equation relating a function to its derivatives.
Separable: one that can be written $g(y)\,dy = f(x)\,dx$.
General solution: the whole family, with its constant.
Particular solution: the one member picked out by an initial condition.
Slope field: a short segment drawn at each point with the slope the equation prescribes there.
A differential equation relates a function to its derivatives, and solving one means finding the functions that satisfy it. Every one has a family of solutions; an initial condition picks a single member.
Separable equations. If the equation can be rearranged so that every $y$ sits with the $dy$ and every $x$ with the $dx$, integrate both sides:
$$\frac{dy}{dx} = f(x)g(y) \;\Longrightarrow\; \int \frac{dy}{g(y)} = \int f(x)\,dx$$
One constant is enough. Then apply the initial condition, and solve for $y$ if an explicit form is wanted.
The exponential model. The most important case is $\dfrac{dy}{dt} = ky$ — the rate proportional to the amount — whose solution is $y = y_0e^{kt}$. Positive $k$ is growth, negative is decay, and a doubling or half-life $T$ gives $k = \dfrac{\ln 2}{T}$ or $\dfrac{\ln \frac12}{T}$. This single equation describes populations, compound interest, radioactive decay and the charge on a capacitor, because all four have the same sentence behind them.
Slope fields. Drawing a short segment at each point with the slope the equation prescribes produces a picture of the solutions before any of them is found. Where the slopes are horizontal there are equilibria; where they steepen the solutions run away. It is worth reading a field before solving, because the field shows what kind of behaviour to expect from an answer.
Another way: picture
Picture a field of iron filings, each one a short line showing the direction to travel. Drop a pin anywhere and follow the filings: the path you trace is a solution. Drop it somewhere else and you trace a different one. The equation is the field; the initial condition is where the pin lands.
Another way: steps
| The sentence | The equation | The behaviour |
|---|---|---|
| Rate proportional to the amount | $y' = ky$ | Exponential; no ceiling |
| Rate proportional to the gap from $M$ | $y' = k(M - y)$ | Approaches $M$ and stops |
| Rate constant | $y' = k$ | A straight line |
Reading which quantity the rate is proportional to is the whole of the modelling. The middle row is Newton's law of cooling, and it is the one people write as the top row by mistake.
Applying the initial condition too early. The constant appears when you integrate. Using $y(0) = 3$ before there is a $C$ to determine achieves nothing and usually loses the condition altogether.
"A constant on each side." Integrating both sides produces two constants, and their difference is a single constant. Carrying both is not wrong, only wasteful — but forgetting that they combine causes people to think a step has gone missing.
Reading a slope field as a graph. It is not the solution; it is the field of directions the solutions follow. Infinitely many curves thread through it, one for each initial condition, and no single one is drawn.
$\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 3$: separate to $y\,dy = x\,dx$.
Variables apart.
Integrate: $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$, so $y^2 = x^2 + 2C$.
One constant.
$y(0) = 3$ gives $9 = 2C$, so $y^2 = x^2 + 9$ and $y = \sqrt{x^2 + 9}$.
Positive root, from the condition.
Half-life 5 years: $e^{5k} = \tfrac{1}{2}$.
Write what the factor must be.
$5k = \ln \tfrac{1}{2} \approx -0.693$, so $k \approx -0.1386$.
Negative, for decay.
After 12 years the fraction left is $e^{-0.1386 \times 12} \approx 0.19$ — under a fifth.
Proportional to size, so $P = 200e^{0.03t}$.
The standard solution.
At $t = 10$: $200e^{0.3} \approx 270$.
Doubling takes $\dfrac{\ln 2}{0.03} \approx 23$ years — and it takes the same 23 years to double again, from any starting point.
Solve $\dfrac{dy}{dx} = 3x y^{-1}$ with $y(0) = 3$, and give $y$ when $x = 3$. (Take $y > 0$.)
answer
Integrating $\dfrac{dy}{dx} = 5$ gives $y = 5x + C$. What does the condition $y(0) = 6$ add?
A population satisfies $\dfrac{dP}{dt} = P$ with $P(0) = 4$ thousand. Write $P(2)$ as a multiple of $e$ raised to a power — what is that power?
answer
A quantity $halves$ every $5$ years under $\dfrac{dQ}{dt} = kQ$. What is $k$, to four decimal places?
answer
A slope field shows horizontal segments all along the $x$-axis, and segments that get steeper the further from it you go — upward above the axis and downward below. Which differential equation is it?
Write $k$ times the right quantity for each. (a) “A colony grows at a rate proportional to its size $P$.” (b) “A cup of coffee at temperature $T$ cools at a rate proportional to how far it is above room temperature, $33$ degrees.”
(a) k times a; (b) k times b
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of solving a separable differential equation in order.
Number the steps in order (write the number in the box):
You can solve separable differential equations. Without looking: at what point in the work does the initial condition get used, and what solves $\dfrac{dy}{dt} = ky$?
9. Your turn: $\dfrac{dP}{dt} = 0.03P$ with $P(0) = 200$, step 3