I2, I3 - Initial conditions and models | AQA A-Level Further Mathematics

A differential equation gives a rate law; an initial condition selects the solution that describes a particular situation. I2 develops this choice of constants, and I3 turns statements about motion and temperature into first-order models. You will interpret the resulting functions, including their units, valid times and limitations.

Selecting a particular solution

A drink starts at $80\,^{\circ}\mathrm{C}$ in a room at $20\,^{\circ}\mathrm{C}$ . If its cooling rate is proportional to its temperature difference from the room, will it ever reach room temperature exactly? We need both a rate law and a starting temperature to answer that question.

For example, a proposed rate law $y'=-0.1(y-20)$ has the general solution $y=20+Ce^{-0.1t}$ . Different values of $C$ describe different starting temperatures; the rate law alone does not choose one. The condition $y(0)=80$ gives $C=60$ . We will interpret that solution in the last part.

Particular solution

A particular solution is a solution with its arbitrary constants fixed by stated conditions. An initial condition specifies a value of the unknown function, or a relevant derivative, at a stated value of the independent variable. Substitute into the solution, not just into one isolated term.

For a first-order linear equation with continuous coefficients on an interval, one value $y(x_0)=y_0$ inside that interval selects exactly one solution. Indeed, two solutions with the same value have difference $C/\mu$ ; at $x_0$ , this forces $C=0$ , so the solutions coincide throughout that interval.

Worked example 1: a condition at zero

Question. Solve $y'+2y=8$ , given $y(0)=7$ .

Plan. Find the general solution using an integrating factor, then substitute the complete initial condition.

\begin{aligned} (e^{2x}y)'&=8e^{2x} && \text{multiply by }e^{2x}\\ e^{2x}y&=4e^{2x}+C && \text{integrate}\\ y&=4+Ce^{-2x}\\ 7&=4+C && \text{use }x=0\\ C&=3 \end{aligned}

Answer. $y=4+3e^{-2x}$ .

Check. At zero, $y=7$ . Also $y'+2y=-6e^{-2x}+8+6e^{-2x}=8$ .

Keep the differential equation and the value $7$ , but move the condition to $x=\ln2$ . The exponential no longer equals one at the specified point.

Worked example 2: a condition away from zero

Question. Solve $y'+2y=8$ , given $y(\ln2)=7$ .

Plan. Use the same general solution, evaluating the exponential at the actual stated point.

\begin{aligned} (e^{2x}y)'&=8e^{2x}\\ y&=4+Ce^{-2x} && \text{integrate and divide}\\ 7&=4+Ce^{-2\ln2}=4+\frac C4\\ C&=12 \end{aligned}

Answer. $y=4+12e^{-2x}$ .

Check. $e^{-2\ln2}=1/4$ gives $y(\ln2)=7$ . Direct substitution gives $y'+2y=-24e^{-2x}+8+24e^{-2x}=8$ .

The second curve has $y(0)=16$ , which is consistent with reaching $7$ later. The same value at different times need not select the same curve.

Watch out

Wrong: $y(\ln2)=7\Rightarrow C=7-4$ . That would use $x=0$ . Substitute the stated independent-variable value into every term before solving for $C$ .

Select a new curve with a condition away from zero.

Formulating rate and motion models

A model starts with a signed statement about change. For an amount, write “rate in minus rate out”. For straight-line motion, choose a positive direction and use $\mathrm{d}v/\mathrm{d}t=\text{acceleration}$ ; a force balance then gives $m\,\mathrm{d}v/\mathrm{d}t=\text{resultant force}$ . The mass-times-acceleration rule is enough here; no additional motion formula is needed.

For a well-mixed tank of fixed volume $V$ , containing amount $q$ , an outflow of $r$ volume units per unit time removes material at rate $rq/V$ . If material enters at rate $a(t)$ , then $q'=a(t)-(r/V)q$ . “Well mixed” is the assumption that makes the exiting concentration equal to $q/V$ .

For cooling in an environment at temperature $u(t)$ , use $y'=-k(y-u)$ , where $k>0$ . Above the environment, $y'<0$ ; below it, $y'>0$ . This sign check is more reliable than interpreting the word “rate” as automatically positive. The units of $k$ are reciprocal time.

Worked example 3: constant driving force

Question. A particle of mass $2\,\mathrm{kg}$ moves in a straight line. The forward force is $6\,\mathrm{N}$ , and resistance is $2v\,\mathrm{N}$ , opposing forward velocity $v\,\mathrm{m}\,\mathrm{s}^{-1}$ . It starts from rest at $t=0$ . Find $v(t)$ and its displacement $s(t)$ from the starting point.

Plan. Form the signed force equation, solve for velocity, then integrate velocity with $s(0)=0$ .

\begin{aligned} 2v'&=6-2v && \text{forward minus resistance}\\ v'+v&=3\\ (e^tv)'&=3e^t && \text{integrating factor }e^t\\ v&=3+Ce^{-t}\\ v(0)=0&\Rightarrow C=-3\\ s&=3t+3e^{-t}+D && \text{integrate velocity}\\ s(0)=0&\Rightarrow D=-3 \end{aligned}

Answer. $v=3-3e^{-t}$ and $s=3t-3+3e^{-t}$ , for $t\ge0$ , in metres per second and metres respectively.

Check. $2v'=6e^{-t}=6-2v$ , $s'=v$ , and both starting values are zero. Velocity tends to $3\,\mathrm{m}\,\mathrm{s}^{-1}$ , where the two forces balance.

Now change only the driving force to one that decreases with time. The same resistance law and starting conditions remain.

Worked example 4: a changing driving force

Question. In Worked example 3, replace the forward force by $(6-2t)\,\mathrm{N}$ , for $0\le t\le3$ . Find $v(t)$ and $s(t)$ on this interval.

Plan. Keep the time-dependent force inside the differential equation and integrate its product with the factor.

\begin{aligned} 2v'&=6-2t-2v\\ (e^tv)'&=(3-t)e^t && \text{normalise and multiply}\\ e^tv&=(4-t)e^t+C && \text{integration by parts}\\ v&=4-t+Ce^{-t}\\ v(0)=0&\Rightarrow C=-4\\ s&=4t-\frac12t^2+4e^{-t}+D\\ s(0)=0&\Rightarrow D=-4 \end{aligned}

The integration step can be checked immediately: $((4-t)e^t)'=(3-t)e^t$ .

Answer. $v=4-t-4e^{-t}$ , $s=4t-\frac12t^2+4e^{-t}-4$ , for $0\le t\le3$ .

Check. $v'+v=(-1+4e^{-t})+(4-t-4e^{-t})=3-t$ ; also $s'=v$ and the two initial conditions hold. Here $v\ge0$ : it rises then falls, with endpoint values $0$ and $1-4e^{-3}>0$ , so the assumed resistance direction is consistent throughout the interval.

The formula may exist for times beyond $3$ , but the stated force model has only been specified up to that time. A mathematical expression does not extend a model's licence to describe the world.

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I2, I3 - Initial conditions and models

Selecting a particular solution

Formulating rate and motion models

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