Differential Equations

A nurse sets up a drip. Medicine flows into the blood at a steady rate, and the body removes it at a rate that grows with the amount already there. Nobody can write a formula for the amount $M(t)$ straight away, but everyone can say how fast it changes. That sentence is a differential equation; this unit teaches you to write one, read its picture, approximate it and solve it.

A differential equation 微分方程 relates a function to its derivatives 导数. Many real situations are described by a rate: “the population grows at a rate proportional to its size” becomes $\dfrac{dP}{dt}=kP$, where $k$ is a constant. Setting up the equation from a verbal description – identifying what changes and what it is proportional 成正比 to – is the first skill.

Worked example: from words to an equation. Read "is proportional to" as "equals a constant $k$ times", and "is inversely proportional to" as "equals $k$ divided by".

sentence	differential equation
The temperature $T$ of a drink changes at a rate proportional to the difference between $T$ and the room temperature $20^{\circ}$C.	$\dfrac{dT}{dt}=k(T-20)$
The depth $h$ of water in a tank decreases at a rate proportional to $\sqrt h$.	$\dfrac{dh}{dt}=-k\sqrt h$, with $k>0$
A rumour spreads at a rate proportional to the product of the number $N$ who have heard it and the number $1000-N$ who have not.	$\dfrac{dN}{dt}=kN(1000-N)$

The word "decreases" gives the minus sign. The last equation is logistic (see the end of this unit).

A solution 解 is a function that satisfies the equation. To verify 验证 a proposed solution, differentiate it and substitute into the equation, checking that both sides agree. A general solution 通解 contains a constant $C$; a particular solution 特解 fixes $C$ from a condition.

Worked example: an unknown linear solution. For what values of $m$ and $b$ is the line $y=mx+b$ a solution of $\dfrac{dy}{dx}=2x-y$?

Substitute: the left side is $m$; the right side is $2x-(mx+b)=(2-m)x-b$. For $m=(2-m)x-b$ to hold for every $x$, the $x$-terms must match ($0=2-m$, so $m=2$) and the constants must match ($m=-b$, so $b=-2$). The solution is $y=2x-2$. Check: $\dfrac{dy}{dx}=2$ and $2x-(2x-2)=2$.

Common mistake. Checking a single point. A solution must make the equation true at every $x$ in its domain.

A slope field 斜率场 draws a short line segment at many points, each with the slope $\dfrac{dy}{dx}$ the equation gives there. It pictures the family of solution curves without solving. To sketch one, evaluate the right-hand side at each grid point and draw a segment of that slope.

Worked example. For $\dfrac{dy}{dx}=x-y$, substitute each point into $x-y$:

point $(x,y)$	$(0,0)$	$(1,0)$	$(0,1)$	$(1,1)$	$(2,1)$	$(-1,1)$
slope $x-y$	$0$	$1$	$-1$	$0$	$1$	$-2$

Every point on the line $y=x$ has slope $0$ (flat segments), and every point on $y=x-1$ has slope $1$.

A solution curve 解曲线 follows the segments like a boat following a current. From a slope field you can sketch the particular solution through a given point, describe long-run behavior, and locate where solutions level off (slopes near zero) – reasoning about solutions purely from the picture.

Reading the field above. The right-hand side depends only on $M$, so every horizontal row of segments is identical. On the line $M=40$ the slope is $0$ at every $t$, so that line is itself a solution: an equilibrium solution 平衡解. Below it solutions rise; above it they fall; none crosses the line.

Exam skill: "Explain why this could not be the slope field." Find one point where the picture disagrees with the equation. For $\dfrac{dH}{dt}=-\dfrac{1}{15}(H-20)$, every segment with $H>20$ must have negative slope; a field showing positive slopes above $H=20$, or slopes that change along a horizontal row, cannot be it. Name the point and both slopes.

Concavity from the equation. Differentiate the equation itself: $\dfrac{d^2M}{dt^2}=-\dfrac14\dfrac{dM}{dt}=-\dfrac1{16}(40-M)$. This is negative for $M<40$, so the solution through $(0,5)$ is concave down.

Euler's method 欧拉方法 approximates a solution numerically by stepping along the slope field. Starting from a known point, take a small step $\Delta x$ and update:

Repeat for each step. Smaller steps give a better approximation 近似值. (This is a BC-only technique.)

Exam skill: be able to carry out two or three Euler steps by hand from a table, and know that Euler's method underestimates 低估 or overestimates 高估 depending on the solution's concavity.

Worked example. Approximate $y(1)$ for $\dfrac{dy}{dx}=x+y$, $y(0)=1$, with step $\Delta x=0.5$. Step 1: slope at $(0,1)$ is $0+1=1$, so $y(0.5)\approx 1+1(0.5)=1.5$. Step 2: slope at $(0.5,1.5)$ is $0.5+1.5=2$, so $y(1)\approx 1.5+2(0.5)=2.5$.

Common mistakes. Forgetting to multiply the slope by $\Delta x$, and working out each slope from the original point instead of the point you have just reached: step 2 uses $(0.5,\,1.5)$.

Over or under? Differentiate the equation: $\dfrac{d^2y}{dx^2}=1+\dfrac{dy}{dx}=1+x+y$, which is positive near these points. The solution is concave up, so each tangent step lies below the curve and $2.5$ is an underestimate. (The exact solution is $y=2e^x-x-1$, so $y(1)=2e-2\approx 3.44$.)

In separation of variables 分离变量法, a separable 可分离 differential equation is written with all the $y$'s on one side and all the $x$'s on the other, then integrated (where the divided factor is nonzero):

This produces the general solution (with a $+C$), the main analytic method for solving differential equations in this course. Also check constant solutions 常数解 with $h(y)=0$ separately; division by $h(y)$ can lose them.

Worked example. Solve $\dfrac{dy}{dx}=xy$ with $y(0)=2$. Separating, $\int\frac{dy}{y}=\int x\,dx$ gives $\ln|y|=\frac{x^2}{2}+C$, so $y=Ae^{x^2/2}$. The condition $y(0)=2$ gives $A=2$, so $y=2e^{x^2/2}$.

Common mistakes. Adding the constant at the end: $y=e^{x^2/2}+C$ is wrong, because the constant enters when you integrate and becomes the multiplier $A=e^{C}$. Trying to separate an equation such as $\dfrac{dy}{dx}=x+y$: a sum cannot be written as $g(x)\,h(y)$.

Worked example (integration by parts inside). Find the particular solution of $\dfrac{dy}{dx}=y\,(x\ln x)$ with $f(1)=4$.

(By parts with $u=\ln x$, $dv=x\,dx$.) At $(1,4)$: $\ln4=0-\tfrac14+C$, so $C=\ln4+\tfrac14$. Since $y>0$,

Worked example (a critical point from the equation). $\dfrac{dH}{dt}=\dfrac12(H-1)\cos\!\left(\dfrac t2\right)$ with $H(0)=4$. Separate: $\displaystyle\int\frac{dH}{H-1}=\int\frac12\cos\!\left(\frac t2\right)dt$, so $\ln|H-1|=\sin\!\left(\dfrac t2\right)+C$. At $t=0$: $\ln3=C$, and

For $0<t<5$, $H>1$, so $\dfrac{dH}{dt}=0$ only when $\cos\!\left(\tfrac t2\right)=0$, at $t=\pi$. The factor $\cos\!\left(\tfrac t2\right)$ changes from positive to negative there, so $\dfrac{dH}{dt}$ does too: $H$ has a relative maximum at $t=\pi$. You can answer this from the differential equation alone, without the solution.

An initial condition 初始条件 (a known point, e.g. $y(0)=5$) pins down the constant $C$. Solve for the general solution, substitute the condition to find $C$, then write the particular solution. Watch the domain 定义域 – a particular solution is valid only on the interval containing the initial point.

Worked example. Solve $\dfrac{dy}{dx}=\dfrac{x}{y}$ with $y(0)=-3$. Separating gives $\dfrac{y^2}{2}=\dfrac{x^2}{2}+C$. Substitute the initial point now: $\tfrac92=C$, so $y^2=x^2+9$. The initial value is negative, so take the negative root:

The positive root fails $y(0)=-3$; choosing the sign from the initial condition is a scored step.

Worked example (domain). $\dfrac{dy}{dx}=xy^2$ with $y(1)=1$: $-\dfrac1y=\dfrac{x^2}{2}+C$ gives $C=-\tfrac32$, so $y=\dfrac{2}{3-x^2}$. The formula breaks at $x=\pm\sqrt3$; the solution lives on the interval that contains $x=1$, so its domain is $-\sqrt3<x<\sqrt3$.

The equation $\dfrac{dy}{dt}=ky$ says the rate of change is proportional to the amount – giving exponential growth or decay 指数增长或衰减. Separating variables yields

with $k>0$ for growth and $k<0$ for decay. This models unrestricted population growth, radioactive decay 放射性衰变, and continuously compounded interest 连续复利.

Worked example: Newton's law of cooling. Coffee at $90^{\circ}$C cools in a room at $20^{\circ}$C by $\dfrac{dT}{dt}=-k\,(T-20)$, with $t$ in minutes. After $5$ minutes it is $70^{\circ}$C. When does it reach $40^{\circ}$C?

The difference $T-20$ decays exponentially from $90-20=70$, so

From $T(5)=70$: $50=70e^{-5k}$, so $k=\dfrac{\ln(7/5)}{5}\approx 0.0673$. For $T=40$: $20=70e^{-kt}$, so $t=\dfrac{\ln 3.5}{k}\approx 18.6$ minutes. Writing $T=90e^{-kt}$ (cooling towards $0^{\circ}$C) is the common error.

Worked example: limited growth. The amount of medicine $M$, in milligrams, satisfies $\dfrac{dM}{dt}=\tfrac14(40-M)$, $t$ in hours, $M(0)=5$.

(a) Separate: $-\ln|40-M|=\dfrac t4+C$, and $C=-\ln35$. Since $M<40$, $M=40-35e^{-t/4}$.

(b) $\displaystyle\lim_{t\to\infty}M(t)=40$: over a long time the amount of medicine in the blood approaches $40$ milligrams. Give the units and the context.

(c) The tangent line at $t=0$ has slope $M'(0)=8.75$, so $M(2)\approx5+8.75(2)=22.5$ mg. Since $\dfrac{d^2M}{dt^2}<0$ (from the slope-field section), the curve is concave down and lies below its tangents: $22.5$ is an overestimate. (Exact: $M(2)\approx18.8$ mg.)

Real growth is limited by resources, so the logistic model 逻辑斯蒂模型 adds a carrying capacity 环境容纳量 $L$:

For $k>0$, $L>0$ and $0<P(0)<L$, growth is nearly exponential when $P$ is small, then the growth rate falls after $P=L/2$ as the population approaches $L$. The states $P=0$ and $P=L$ are equilibrium solutions; a population initially at zero stays zero. Key BC facts: the population levels off at $L$ ($\lim_{t\to\infty}P=L$), and it grows fastest when $P=\tfrac{L}{2}$ (the inflection point of the S-shaped curve). You are expected to read $L$ and the fastest-growth value directly from the equation. If $P(0)>L$, the population instead decreases towards $L$, so $\lim_{t\to\infty}P=L$ still holds.

Worked example. For $\dfrac{dP}{dt}=0.05\,P\!\left(1-\dfrac{P}{2000}\right)$, the carrying capacity is $L=2000$ (the population levels off there), and growth is fastest when $P=\dfrac{L}{2}=1000$ – both read straight off the equation, no solving needed.

Common mistakes. Saying growth is fastest when $P=L$: at $P=L$ the growth rate is $0$. Giving a time when the question asks for the population at which growth is fastest ($P=1000$).

Why $\tfrac L2$? Write $\dfrac{dP}{dt}=kP-\dfrac kLP^2$. Differentiate with respect to $t$: $\dfrac{d^2P}{dt^2}=\left(k-\dfrac{2k}{L}P\right)\dfrac{dP}{dt}$. For $0<P<L$, $\dfrac{dP}{dt}>0$, so $\dfrac{d^2P}{dt^2}$ changes sign where $k-\dfrac{2k}{L}P=0$, at $P=\dfrac L2$. That is where the growth rate is largest.

Worked example (reading the logistic form). $\dfrac{dP}{dt}=\dfrac{P(600-P)}{300}$ with $P(0)=800$. Rewrite: $\dfrac{dP}{dt}=2P\left(1-\dfrac{P}{600}\right)$, so $k=2$ and $L=600$. Because $P(0)>600$, $\dfrac{dP}{dt}<0$: the population decreases towards $600$, and $\displaystyle\lim_{t\to\infty}P(t)=600$. It never passes through $\tfrac L2=300$, so it has no point of fastest growth.

• Solve a separable equation by getting all $y$ on one side and all $x$ on the other, then integrating both sides (add $+C$ once).
• Use the initial condition to find $C$ (a particular solution).
• Sketch or read a slope field: the little segments show $\tfrac{dy}{dx}$ at each point, and a solution curve follows them.
• Recognise exponential models $\tfrac{dy}{dt}=ky\Rightarrow y=Ce^{kt}$ (growth/decay).
• A differential equation gives the slope — you must integrate to recover the function.
• Logistic $\tfrac{dP}{dt}=kP\left(1-\tfrac PL\right)$: the limit is $L$ (for $P(0)>0$) and growth is fastest at $P=\tfrac L2$.

Practice.

1. Verify that $y=3e^{2x}-1$ is a solution of $\dfrac{dy}{dx}=2y+2$.
2. For $\dfrac{dy}{dx}=y-x$, use Euler's method with two steps of size $0.5$ from $y(0)=2$ to estimate $y(1)$.
3. Find the particular solution of $\dfrac{dy}{dx}=\dfrac{x^2}{y}$ with $y(0)=-2$.
4. $\dfrac{dA}{dt}=\dfrac{12-A}{3}$, $A(0)=0$, with $A$ in grams and $t$ in minutes. Find $A(t)$ and interpret $\displaystyle\lim_{t\to\infty}A(t)$.
5. A population satisfies $\dfrac{dP}{dt}=0.4P-0.0002P^2$ with $P(0)=300$. Find $\displaystyle\lim_{t\to\infty}P(t)$ and the population when it grows fastest.
6. $\dfrac{dy}{dt}=0.5(y-10)$ with $y(0)=4$. Find $\dfrac{d^2y}{dt^2}$ in terms of $y$, use the tangent line at $t=0$ to estimate $y(1)$, and decide whether the estimate is too big or too small.

Answers.

1. $\dfrac{dy}{dx}=6e^{2x}$ and $2y+2=6e^{2x}$; the sides agree for all $x$.
2. Slope at $(0,2)$ is $2$, so $y(0.5)\approx2+0.5(2)=3$. Slope at $(0.5,3)$ is $2.5$, so $y(1)\approx3+0.5(2.5)=4.25$.
3. $\dfrac{y^2}{2}=\dfrac{x^3}{3}+2$, and $y(0)<0$, so $y=-\sqrt{\dfrac{2x^3}{3}+4}$ for $x>-\sqrt[3]{6}$.
4. $A=12-12e^{-t/3}$; the limit is $12$: over a long time the amount approaches $12$ grams.
5. $0.4P\left(1-\dfrac{P}{2000}\right)$: $L=2000$, so the limit is $2000$; fastest growth at $P=1000$.
6. $\dfrac{d^2y}{dt^2}=0.25(y-10)$. $y'(0)=-3$, so $y(1)\approx1$. For $y<10$ the solution is concave down, so the estimate is too big (exact $10-6e^{0.5}\approx0.11$).