Unit 7: Differential Equations

17 terms. Say what each one means, then open the card to check.

• Differential equation

  An equation that links an unknown function to one or more of its derivatives. Solving it means finding the function, not a single number.

  Example. $\frac{dV}{dt} = -0.2V$ says the volume of water in a leaking jug shrinks at a rate equal to 0.2 times the volume left.

• Solution of a differential equation

  A function that makes the equation true when you substitute it and its derivative. Check one by differentiating and comparing both sides.

  Example. $y = x^2$ solves $\frac{dy}{dx} = \frac{2y}{x}$ for $x \ne 0$: the left side is $2x$ and the right side is $\frac{2x^2}{x} = 2x$.

• Proportional rate

  A rate equal to a constant times some quantity. "The rate of change of y is proportional to Q" is written $\frac{dy}{dt} = kQ$.

  Example. If water evaporates from a pond at a rate proportional to its surface area A, then $\frac{dV}{dt} = -kA$ with $k > 0$.

• Initial condition

  A known value of the solution at one input, such as $y(0) = 5$. It is used to find the constant of integration.

  Example. A tank that holds 300 liters when a valve opens gives the initial condition $V(0) = 300$.

• Second derivative from the equation

  Differentiate the right side of $\frac{dy}{dx}$ with respect to x, using the chain rule on y, then substitute $\frac{dy}{dx}$ back in. Its sign gives concavity.

  Example. For $\frac{dy}{dx} = y^2$, $\frac{d^2y}{dx^2} = 2y\frac{dy}{dx} = 2y^3$, so solutions with $y > 0$ are concave up.

• Newton’s law of cooling

  An object's temperature changes at a rate proportional to the difference between its temperature and the surroundings: $\frac{dT}{dt} = -k(T - A)$ with $k > 0$.

  Example. A loaf at 82 degrees in a 22-degree kitchen with $\frac{dT}{dt} = -0.05(T - 22)$ cools at 3 degrees per minute at first.

• Slope field

  A grid of short segments, each drawn with the slope the differential equation gives at that point. It shows the shape of the solutions without solving.

  Example. For $\frac{dy}{dx} = x + y$, the segment at $(1, 2)$ has slope 3 and the segment at $(2, -2)$ is flat.

• Solution curve

  The graph of one particular solution. In a slope field it runs along the segments, tangent to each one it passes through.

  Example. For $\frac{dy}{dx} = -\frac{x}{y}$, the solution curve through $(0, 3)$ is the upper half of the circle $x^2 + y^2 = 9$.

• Equilibrium solution

  A constant solution, found where the derivative is zero for every input. In a slope field it shows as a whole row of horizontal segments.

  Example. For $\frac{dy}{dt} = 3(10 - y)$, the constant function $y = 10$ is an equilibrium solution.

• Separation of variables

  Rewrite the equation with every y on the side with dy and every x on the side with dx, then integrate both sides. It works when the right side is a function of x times a function of y.

  Example. $\frac{dy}{dx} = x^2y$ separates to $\frac{dy}{y} = x^2\,dx$, so $\ln|y| = \frac{x^3}{3} + C$.

• General solution

  The family of all solutions, written with an arbitrary constant such as C. Each value of the constant gives a different curve.

  Example. $y = Ce^{3x}$ is the general solution of $\frac{dy}{dx} = 3y$; $C = 2$ and $C = -1$ give two of its curves.

• Particular solution

  The one solution that also meets a given condition, such as passing through a stated point. The condition fixes the constant.

  Example. For $\frac{dy}{dx} = 3y$ with $y(0) = 4$, the particular solution is $y = 4e^{3x}$.

• Constant of integration

  The C added after integrating both sides. One C is enough, and it is easiest to find from the initial condition right after integrating.

  Example. From $\ln|y| = 2x + C$ and $y(0) = 1$, $C = \ln 1 = 0$, so $y = e^{2x}$.

• Exponential growth and decay

  When a quantity changes at a rate proportional to itself, $\frac{dy}{dt} = ky$, the solution is $y = y_0e^{kt}$. It grows if $k > 0$ and decays if $k < 0$.

  Example. A fictional cell count with $\frac{dB}{dt} = 0.9B$ and 40 cells at the start follows $B = 40e^{0.9t}$.

• Growth constant k

  The constant in $\frac{dy}{dt} = ky$: the rate of change per unit of the quantity, per unit of time. It is not the factor y is multiplied by each unit of time.

  Example. With $k = 0.1$ per year, y is multiplied by $e^{0.1} \approx 1.105$ each year, not by 1.1.

• Doubling time

  The time a growing exponential quantity takes to double. From $e^{kt} = 2$ it is $\frac{\ln 2}{k}$, the same from any starting amount.

  Example. With $k = 0.035$ per year, the doubling time is $\frac{\ln 2}{0.035} \approx 19.8$ years.

• Half-life

  The time a decaying exponential quantity takes to fall to half. If $\frac{dy}{dt} = -ry$ with $r > 0$, the half-life is $\frac{\ln 2}{r}$.

  Example. A fictional isotope with a half-life of 8 days has $r = \frac{\ln 2}{8} \approx 0.0866$ per day.