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State, rate, and solution

Chapter 1 - Read before you solve

An ordinary differential equation is a sentence about change. It names a state, gives a rule for the state's rate, and asks for a solution function whose motion obeys that rule.

By the end of this chapter, you should be able to:

  • identify the independent variable, state, rate, and parameters;
  • translate a verbal rate law into an equation;
  • distinguish a differential equation from one of its solutions; and
  • verify a proposed solution by substitution and by behavior.

The three nouns

Suppose (T(t)) measures a cup of coffee's temperature in degrees Celsius, while (t) measures minutes. Then:

  • (T) is the state, with units °C;
  • (T') is the rate, with units °C/min;
  • room temperature, say (22) °C, and a cooling constant (k) are parameters;
  • (T'=-k(T-22)) is the differential equation; and
  • (T(t)=22+68e^{-kt}) is the solution that starts at (90) °C.

The equation is a local rule: it tells the slope at the current state. A solution collects those local instructions into an entire curve.

Say the equation aloud

"The temperature changes at a rate proportional to its signed distance from room temperature, directed back toward the room." If you cannot say an equation in words, the symbols have not become usable language yet.

Units expose meaning

Every term added to another term must have the same units. In

the left side has units °C/min. The factor ((T-22)) has units °C, so (k) must have units (1/\text{min}).

This gives a fast error detector. The equation (T'=-(T-22)^2) cannot use a dimensionless proportionality constant: its right side would have units °C² rather than °C/min.

Unit check. In \(P'=rP(1-P/K)\), population \(P\) is measured in cells and time in hours. What are the units of \(r\)?

Worked example: read a model from a story

Medicine leaves the bloodstreamfully worked

A patient starts with 120 mg of a medicine in the bloodstream. The body removes 18% of the current amount per hour. Build an initial-value problem and predict the graph before solving.

  1. Name the state. Let \(A(t)\) be the amount in milligrams after \(t\) hours. Then \(A'\) has units mg/hour.
  2. Translate the rate phrase. "18% of the current amount per hour" gives magnitude \(0.18A\). "Removes" supplies the negative sign: \(A'=-0.18A\).
  3. Attach the initial condition. "Starts with 120 mg" means \(A(0)=120\).
  4. Predict before solving. When \(A>0\), the rate is negative, so the graph decreases. As \(A\) shrinks, \(|A'|\) shrinks, so the curve flattens toward zero.
  5. State the model. \(A'=-0.18A,\quad A(0)=120\). Chapter 4 will solve it; for now, its behavior already says a great deal.

Fade the support

I do

"Grows by 7% of its current size per year" becomes \(P'=0.07P\).

We do

"Loses 4 grams plus 3% of its current mass per day" becomes \(M'=\underline{\hspace{3em}}\).

Check the two rate contributions

\(M'=-4-0.03M\).

You do

A tank receives 5 L/min of clean water and drains 5 L/min. If \(S\) is grams of salt in a 100 L tank, write the salt rate.

One cue

Rate in minus rate out. Clean inflow carries no salt.

The last model is (S'=0-(5\text{ L/min})(S/100\text{ g/L})=-S/20) g/min.

A solution is a function, not a final number

Students often carry an algebra habit into differential equations: "solve" should produce a number. Here it usually produces a function.

For (y'=2y):

  • the differential equation names a family of possible motions;
  • (y=Ce^{2t}) is the general solution family;
  • (y(0)=3) selects the particular solution (y=3e^{2t}); and
  • (y(4)=3e^8) is one value of that solution.
A family of exponential solution curves Several curves satisfy the same differential equation, while one initial condition selects a single curve. initial condition selects one curve time state
One rate law can admit many solution curves. An initial condition chooses one.

Verify in three passes

To test the claim that (y=4e^{-3t}) solves (y'=-3y) with (y(0)=4):

  1. Differentiate: (y'=-12e^{-3t}).
  2. Substitute: (-3y=-3(4e^{-3t})=-12e^{-3t}=y').
  3. Check the condition and behavior: (y(0)=4), the curve stays positive, decreases, and flattens. That matches a negative rate whose magnitude shrinks with (y).

Symbolic substitution proves the equation. Behavior checks catch sign and interpretation errors that algebra can hide.

Error clinic: checking only the initial value

A student claims \(y=4-3t\) solves \(y'=-3y\) because \(y(0)=4\). The condition is necessary, but it is not enough. Here \(y'=-3\), while \(-3y=-12+9t\); they disagree except at one accidental time.

Practice

  1. Practice - translation

    1.1 A pond contains \(W(t)\) cubic meters of water. Rain adds 12 m³/hour, while leakage removes 4% of the current volume per hour. Write the differential equation and state the units of every term.

    Detailed solution

  2. Practice - verification

    1.2 Verify whether \(y=2+3e^{-t}\) solves \(y'=2-y\) and \(y(0)=5\). Use all three passes.

    Detailed solution

  3. Integrated - representation

    1.3 For \(N'=0.4N(1-N/50)\), decide whether the state initially rises or falls when \(N(0)=20\), \(50\), and \(80\). Do not solve.

    Detailed solution

  4. Challenge - modeling

    1.4 A 200 L tank initially holds 10 kg of dissolved salt. Brine containing 0.08 kg/L enters at 3 L/min; well-mixed solution leaves at 3 L/min. Let \(S(t)\) be kilograms of salt. Build the initial-value problem and identify the long-run amount suggested by a zero rate.

    Detailed solution

Retrieve before you leave

Close the page or cover the text. Say these answers aloud:

  1. What are the four objects you name when reading a model?
  2. Why does a differential equation usually have a family of solutions?
  3. What three passes verify a proposed solution?

Next: See the field of slopes before choosing any solution method.