H2 Math Differential Equations: Study Guide

JC · H2 Mathematics · Study guide

Form and solve exponential, input–output and nonlinear models; handle changing volume, initial conditions, equilibria, limiting values and three forms of given substitution.

Concepts · worked methods · graphs10 min read · practice linked below

Form the rate equation, then solve it

Name the changing quantity and its units. “Proportional to” introduces a constant; “remains constant” gives zero rate of change. For an amount in a system, use rate in minus rate out. Separate variables only after checking any constant solutions that division might remove.

Separation of variables and initial conditions

Calibrate exponential growth from two observations

A sample initially has mass 100 g. Its growth rate is proportional to its mass, and after one day its mass is 110 g. Find its mass $B$ after $t$ days and the time taken to reach 150 g.

\begin{aligned}\frac{dB}{dt}&=kB,\quad k>0,\\\int\frac1B\,dB&=\int k\,dt,\\\ln B&=kt+C,\\B&=Ae^{kt}.\end{aligned}

Use the two observations separately:

\begin{aligned}B(0)=100&\implies A=100,\\B(1)=110&\implies e^k=\frac{11}{10},\\k&=\ln\frac{11}{10},\\B&=100\left(\frac{11}{10}\right)^t.\end{aligned}

\begin{aligned}150&=100(1.1)^t,\\t&=\frac{\ln1.5}{\ln1.1},\\&\approx4.254\text{ days}.\end{aligned}

The model predicts unlimited growth. That is a statement about the model; growth may not remain proportional indefinitely in a real sample.

A time-dependent growth rate can give a finite limit

A population model satisfies $N'=kN/(1+t^2)$ , with $N(0)=10$ and $N(1)=20$ . Here N is measured in thousands and t in years. Find N and its long-term value.

\begin{aligned}\int\frac1N\,dN&=k\int\frac1{1+t^2}\,dt,\\\ln N&=k\tan^{-1}t+C,\\N(0)=10&\implies C=\ln10,\\N&=10e^{k\tan^{-1}t}.\end{aligned}

\begin{aligned}20&=10e^{k\pi/4},\\k&=\frac{4\ln2}{\pi}. \end{aligned}

As time increases, inverse tangent approaches π/2:

\begin{aligned}\lim_{t\to\infty}N&=10e^{(4\ln2/\pi)(\pi/2)},\\&=10e^{2\ln2}=40.\end{aligned}

The population approaches 40 thousand from below. It is increasing at every finite time, yet bounded. Do not assume every positive proportional-growth model is exponential in t.

Forming the differential equation: input minus output

An extraction system and its steady-state

Particles enter a workshop at 0.32 mg/min. An extractor removes particles at a rate proportional to the mass $V$ currently in the air. When $V=18.75$ mg, the net increase is 0.12 mg/min. Initially the air contains no particles. Find V, and decide whether it stays below 32 mg.

\begin{aligned}\frac{dV}{dt}&=0.32-kV,\\0.12&=0.32-18.75k,\\k&=\frac{0.20}{18.75}=\frac4{375},\\\frac{dV}{dt}&=\frac4{375}(30-V).\end{aligned}

At steady-state, input balances output, so V = 30 is a constant solution. For the initially clean room, V is below 30 and separation gives

\begin{aligned}\int\frac{dV}{30-V}&=\int\frac4{375}\,dt,\\-\ln|30-V|&=\frac{4t}{375}+C,\\30-V&=Ae^{-4t/375},\\V(0)=0&\implies A=30,\\V&=30(1-e^{-4t/375}).\end{aligned}

For $t\geq0$ , $0\leq V<30<32$ , so this model meets the recommendation. The value 30 is a limiting value, never reached at finite time from this initial state.

If the extractor removes particles at half its original rate, keep the input unchanged:

\begin{aligned}\frac{dV}{dt}&=0.32-\frac2{375}V,\\\frac{dV}{dt}=0&\implies V=60.\end{aligned}

The new steady-state is 60 mg, so the weakened extractor cannot keep an initially clean room below 32 mg indefinitely.

Initial value and limiting value

For the extraction model, the solution for an arbitrary initial mass $V_0$ is $V=30+(V_0-30)e^{-4t/375}$ . The input and extractor are unchanged. Move the initial value to see which side of 30 the solution approaches from.

V₀ = 0 mg: V increases towards 30 mg. At t = 100 min, V ≈ 19.68 mg.

A changing tank volume: distinguish amount and concentration

A tank initially contains 100 L of pure water. Brine containing 0.2 kg of salt per litre enters at 5 L/min. A well-stirred mixture leaves at 3 L/min. Find the salt concentration $C$ at time t, assuming the tank has enough capacity.

The volume is $W=100+2t$ L. Let Q be salt mass in kg. Salt enters at $0.2(5)=1$ kg/min; it leaves at concentration Q/W times 3 L/min.

\begin{aligned}\frac{dQ}{dt}&=1-\frac{3Q}{W},\\Q&=WC,\\\frac{dQ}{dt}&=W\frac{dC}{dt}+2C.\end{aligned}

Equate the two expressions for the salt-mass rate:

\begin{aligned}W\frac{dC}{dt}+2C&=1-3C,\\\frac{dC}{dt}&=\frac{1-5C}{100+2t},\\\int\frac{dC}{1-5C}&=\int\frac{dt}{100+2t},\\-\frac15\ln|1-5C|&=\frac12\ln W+K.\end{aligned}

Initially C = 0. On the branch below 0.2 kg/L, the logarithm argument is positive:

\begin{aligned}\ln(1-5C)&=-\frac52\ln\left(\frac W{100}\right),\\1-5C&=\left(\frac{100}{W}\right)^{5/2}.\end{aligned}

\begin{aligned}C&=\frac15\left[1-\left(\frac{100}{W}\right)^{5/2}\right],\\W&=100+2t.\end{aligned}

Thus $C\to0.2$ kg/L. The mass of salt does not approach 0.2 kg: the volume is growing, so the mass continues to grow. Equal inflow and outflow would instead make W constant.

A nonlinear model and its equilibrium solutions

An initial value above the limiting value

An inventory model is $x'=3x-x^2/3$ , where x is kg of stock and t is in days. Initially x = 13.5 kg. Solve and interpret the long-term behaviour.

Factor the rate as $x'=x(9-x)/3$ . The constant solutions are $x=0$ and $x=9$ . For this initial condition, x > 9, so it is decreasing.

\begin{aligned}\int\frac{dx}{x(9-x)}&=\int\frac13\,dt,\\\frac1{x(9-x)}&=\frac19\left(\frac1x+\frac1{9-x}\right),\\\ln\left|\frac{x}{9-x}\right|&=3t+C,\\\frac{x}{9-x}&=Ae^{3t}.\end{aligned}

The constant A may be negative: the modulus is removed with a signed constant.

\begin{aligned}x&=\frac{9Ae^{3t}}{1+Ae^{3t}},\\13.5&=\frac{9A}{1+A},\\A&=-3,\\x&=\frac{27}{3-e^{-3t}}.\end{aligned}

For $t\geq0$ , the denominator is at least 2, so the solution has no future pole. Since $e^{-3t}\to0$ , x decreases towards 9 kg from above. It does not cross the equilibrium value. Stock remains positive in this model.

If the initial amount were exactly 9, use the constant solution directly: separation divided by 9 − x, so its derivation did not include that case.

Given substitutions: differentiate the substitution first

Recognise a product derivative

For $0<x<\pi/2$ , $y>0$ , solve $(2xyy'+y^2)\cos x=1/(xy^2)$ using $u=xy^2$ .

\begin{aligned}\frac{du}{dx}&=y^2+2xy\frac{dy}{dx},\\\frac{du}{dx}\cos x&=\frac1u,\\u\frac{du}{dx}&=\sec x,\\\int u\,du&=\int\sec x\,dx,\\\frac12u^2&=\ln(\sec x+\tan x)+K,\\x^2y^4&=2\ln(\sec x+\tan x)+C.\end{aligned}

The log argument is positive on the stated x interval. The right side must also be positive because x and y are positive. The displayed implicit equation is the general solution on intervals satisfying those conditions.

For u = y/x, remember the product rule

Solve $2xyy'=x^2-y^2$ , where $x>0$ , $y<0$ , using $u=y/x$ .

\begin{aligned}y&=ux,\\y'&=u+xu',\\2u(u+xu')&=1-u^2,\\2uxu'&=1-3u^2.\end{aligned}

Before dividing by $1-3u^2$ , check the constant-u case. The negative branch $u=-1/\sqrt3$ gives the valid straight-line solution $y=-x/\sqrt3$ . For the remaining cases:

\begin{aligned}\int\frac{2u}{1-3u^2}\,du&=\int\frac{dx}{x},\\-\frac13\ln|1-3u^2|&=\ln x+K,\\1-3u^2&=\frac A{x^3},\\y^2&=\frac{x^2}{3}\left(1-\frac A{x^3}\right).\end{aligned}

Choose the negative square root because y < 0:

\begin{aligned}y&=-\frac{x}{\sqrt3}\sqrt{1-\frac A{x^3}}.\end{aligned}

The domain must satisfy $x>0$ and $1-A/x^3>0$ . Allowing $A=0$ restores the straight-line solution found before division. A strict inequality is required because y cannot be zero.

A substitution can reduce the order

Given $xy''+y'=(\ln x)/x$ , $x>0$ , and $y'(1)=1$ , use $z=xy'$ to find y′. Then find the tangent at $(e,7/6)$ .

\begin{aligned}\frac{dz}{dx}&=xy''+y',\\\frac{dz}{dx}&=\frac{\ln x}{x},\\z&=\frac12(\ln x)^2+C,\\y'(1)=1&\implies z(1)=1,\\C&=1,\\y'&=\frac{(\ln x)^2}{2x}+\frac1x.\end{aligned}

Evaluate the derivative at x = e, then use the given point:

\begin{aligned}y'(e)&=\frac3{2e},\\y-\frac76&=\frac3{2e}(x-e),\\y&=\frac3{2e}x-\frac13.\end{aligned}

There is no need to integrate again to find y when only its tangent is requested.

Interpret the solution within its model

State the units, the time interval and the limiting quantity. Check an initial condition in the formula and differentiate it to check the original equation. An equilibrium may be approached without being reached. A zero rate at one instant alone does not mean the whole solution is constant; a solution starting at an equilibrium of these smooth autonomous models stays there.

Continue with topic questions

Differential equations practice and worked solutions