← Unit 7

Unit 7 Cheat Sheet

Every formula in Differential Equations — and, more importantly, when to reach for which. The four ways to handle a \(\dfrac{dy}{dx}\) (a picture, a step-by-step march, the exact function, a named model) look interchangeable until a question demands a specific one, so each block ends with the decision that picks the right tool — and the two skills that decide most exam points, translating the words into an equation and reading behavior without solving, get their own panels at the bottom.
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Equations of Change 7.1 – 7.2

A differential equation is a rule for the rate \(\dfrac{dy}{dx}\) written in terms of \(x\), \(y\), or both — a local instruction, not a finished formula for \(y\).

rate proportional to amount
rate proportional to a difference (Newton's law)
verify a candidate solution
concavity — differentiate the equation again

Verifying is not solving. Compute \(y'\) from the candidate, substitute it and \(y\) into the equation, and check both sides agree — never integrate.

Second derivative, concretely: if \(y'=x-y\) then \(y''=1-y'=1-(x-y)\). Differentiate the right side and substitute \(\dfrac{dy}{dx}\) back in — that sign tells you concavity.

Translate "is proportional to…"
the amount\(\dfrac{dy}{dt}=ky\)
a difference\(\dfrac{dy}{dt}=k(y-M)\) — approaches \(M\)
a product\(\dfrac{dP}{dt}=kP(M-P)\) — logistic
trap
  • Concavity comes from the equation, not a guess — differentiate \(\dfrac{dy}{dx}\) and substitute, don't read it off a sketch.
  • Equilibrium solutions are the constant \(y\) where \(\dfrac{dy}{dx}=0\); the sign of \(\dfrac{dy}{dx}\) elsewhere says increasing or decreasing.

Slope Fields 7.3 – 7.4

A grid of tiny segments. At each point \((x,y)\) the segment's slope is just \(\dfrac{dy}{dx}\) evaluated there — no solving required.

slope of the segment at a point
where segments lie flat (equilibrium)

Read the dependence: if \(\dfrac{dy}{dx}\) depends only on \(x\), every column looks identical; only on \(y\), every row looks identical; on both, the field changes everywhere.

Match a field to an equation
flat segmentssolve \(\dfrac{dy}{dx}=0\) — those locations must match the field's horizontal marks
columns repeatthe equation is a function of \(x\) alone
rows repeata function of \(y\) alone
still unsureplug one easy point (like the origin) and compare the slope
trap
  • A field shows slope, not height. To sketch a solution, start at the given point and follow the segments like a current.
  • Curves of equal slope (isoclines) are where \(\dfrac{dy}{dx}=c\) — handy for matching, but they are not solution curves.

Euler's Method 7.5 · BC

When there's no formula for \(y\), march forward in steps of size \(h\), each step following the tangent line at the point you're standing on.

update the y-value
update the x-value

One line to remember: new \(y\) = old \(y\) + (slope at the old point)\(\,\times\,h\). The slope is \(\dfrac{dy}{dx}\) at where you are, not where you're going.

Is the estimate too big or too small?
concave uptangents sit below the curve ⇒ Euler underestimates
concave downtangents sit above the curve ⇒ Euler overestimates

Accuracy: a smaller \(h\) (more steps) shrinks the error — roughly in proportion to \(h\). Get concavity from \(\dfrac{d^2y}{dx^2}\) (differentiate the equation).

trap
  • Update both \(x\) and \(y\) every step — a wrong \(x\) feeds a wrong next slope.
  • Use the current slope. Recomputing the slope at the new point first is a different (better) method, not Euler.

Separation of Variables 7.6 – 7.7

The one technique for finding the actual function: get the \(y\)'s with \(dy\), the \(x\)'s with \(dx\), integrate both sides, then solve.

separate, then integrate both sides
the exponential pattern you'll meet most

The \(+C\) goes in the moment you integrate — before you solve for \(y\) or exponentiate. Then use the initial condition to pin down \(C\).

Is it separable?
yesyou can write \(\dfrac{dy}{dx}=g(x)\,h(y)\) — a function of \(x\) times a function of \(y\)
thendivide by \(h(y)\), multiply by \(dx\), integrate each side
trap
  • Apply the initial condition after integrating, with the \(+C\) present — solving for \(C\) before integrating loses it.
  • Exponentiate cleanly: \(e^{\,x^2+C}=e^{C}e^{\,x^2}\), and \(e^{C}\) is just a new positive constant; the \(\pm\) from a square root is fixed by the initial point.

Exponential & Logistic 7.8 – 7.9 · BC

Two named models worth recognizing on sight — one grows without bound, the other levels off at a ceiling.

exponential — growth or decay
doubling time & half-life
logistic equation (BC)
capacity & where growth is fastest
Exponential or logistic?
∝ amountexponential — unbounded; \(y=y_0e^{kt}\)
∝ productlogistic — levels off at \(L\); fastest at \(L/2\)
long run\(P\to L\) from below if \(P_0\lt L\), from above if \(P_0\gt L\)
trap
  • Capacity is \(a/b\) when the equation reads \(\dfrac{dP}{dt}=aP-bP^2\) — not \(a\), not \(b\).
  • Fastest growth is at half capacity \((L/2)\), the inflection point of \(P\) vs \(t\) — not at \(L\).
  • You needn't solve the logistic equation to find the capacity or the max-rate point — they come straight from the coefficients.

Behavior without solving the shortcut

A surprising share of questions never need the solution — they're answered straight from the equation.

Ask the equation directly
increasing?sign of \(\dfrac{dy}{dx}\) — positive rises, negative falls
concave?sign of \(\dfrac{d^2y}{dx^2}\) (differentiate the equation, substitute)
steady states?solve \(\dfrac{dy}{dx}=0\) for the equilibrium values
long-run?trace toward the stable equilibrium the field flows into

Why it matters: "is the estimate an over- or under-estimate," "is the function concave up at this point," "what value does the solution approach" are all concavity-and-sign questions — fast points if you go to the equation instead of hunting for a formula.

Translate the words into a differential equation the modeling skill

Most modeling points are won or lost in the first line. Map the English phrase straight to a rate.

proportional to the amount
\(\dfrac{dQ}{dt}=kQ\)
unbounded growth/decay — the exponential model.
proportional to the difference
\(\dfrac{dT}{dt}=k(T-M)\)
Newton's heating/cooling — temperature eases toward the surroundings \(M\).
proportional to the product
\(\dfrac{dP}{dt}=kP(M-P)\)
present amount times what's left — logistic; ceiling \(M\).
inversely proportional
\(\dfrac{dy}{dt}=\dfrac{k}{y}\)
rate shrinks as the quantity grows.
"rate of change of …"
\(\dfrac{d(\cdot)}{dt}\)
the subject of that phrase is the thing you differentiate; everything after "is" defines the right side.
read a logistic graph
capacity \(=\) the asymptote
the leveling-off value is \(L\); for \(aP-bP^2\) that's \(a/b\), which picks the right equation.

Then attach the data: an initial value gives \(y_0\) (or the constant \(C\)); a second data point or a known rate solves for \(k\). Units on the rate (people per year, °F per minute) confirm you differentiated the right quantity.

Which tool does the question want? the whole-unit decision

Every Unit 7 problem is one of four asks. Match the wording, then run the matching machinery.

"describe / sketch the behavior"
slope field + sign & concavity
plot or read segments; equilibria where \(\dfrac{dy}{dx}=0\); concavity from the second derivative. No solving.
"approximate the value" (no formula)
Euler's method · BC
step with \(y_{n+1}=y_n+h\,f\); judge over/under from concavity.
"find the function / particular solution"
separate & integrate, then use the IC
\(\int h(y)^{-1}dy=\int g(x)\,dx+C\); solve for \(y\), pin \(C\) with the initial point.
"growth / decay / population" word problem
name the model
exponential \(ky\), Newton \(k(y-M)\), or logistic — read capacity \(=a/b\) and fastest growth at \(L/2\).
Quick tells in the prompt
"approximate"Euler — and check the over/under-estimate
"find y =" / "solution"separation of variables with the initial condition
"carrying capacity / levels off"logistic — capacity and the \(L/2\) inflection
"slope field / sketch / which graph"reason from segments; never solve

The reliable move: read what the verb is asking before touching algebra. "Approximate," "sketch," "find," and "how large does it get" each point at a different one of the four — and choosing right is most of the credit.

dy⁄dx

Unit 7 is a single idea seen four ways: a differential equation is a local rule for change — and you can meet it as a picture (the slope field), a step-by-step march (Euler), the exact function (separation of variables), or a named model (exponential and logistic). The rule never changes; only how much detail the question demands. Read the verb, pick the view, and the calculus you already know finishes it.

Test yourself on the exam set →
Equations of Change Slope Fields Euler's Method Separation Growth Models