Form differential models, verify solutions and read direction fields

Use the existing calculus foundations

Methods derivatives and senior 4A–5B provide differentiation and integration. This section formulates and verifies differential equations; 6B will solve separable and prescribed-second-derivative equations. A proposed solution must satisfy a rate identity on an interval, not just one numerical check

Name the independent and dependent variables

In y′=f(x,y), x is the independent variable and y(x) the unknown function. The equation relates a function to its derivative. An initial condition y(x0)=y0 chooses a point the solution must pass through. Record units and the domain before interpreting the model

Read order without confusing it with degree

The highest derivative gives the order. y′=y²+x is first order despite the square on y; y″=x is second order. A second-order equation usually needs two independent conditions. Order does not by itself prove existence or uniqueness

Use the equation to find a tangent

For y′=x-2y+3 at (1,3), slope m=1-6+3=-2. The local tangent is y-3=-2(x-1), or y=-2x+5. The tangent approximates the solution near that point; it is not generally the whole solution

For y′=x−2y+3, give the tangent slope at (1,3)

The tangent y=−2x+n passes through (1,3). Give n

An isocline collects equal prescribed slopes

For y′=x-2y+3, points with prescribed slope m satisfy y=(x+3-m)/2. At x=1 the slope is 2 when y=1 and zero when y=2. An isocline describes local tangent directions; it need not be a solution curve

At x=1, what y-coordinate gives slope 2 for y′=x-2y+3?

Verify the equation and initial condition separately

Differentiate the candidate, substitute it into the original right-hand side, and simplify their difference. The residual must be identically zero throughout a connected interval where the candidate is differentiable and the equation defined. Then check every initial condition

For y′=2x+3,y(0)=1, which statement describes y=x²+3x+4?

A single matching point is not an identity

The candidate y=x² has derivative 2x. For the equation y′=2x+x², its residual is -x². That residual is zero at x=0 only; agreement there does not make the candidate a solution on an interval

Verify a translated exponential

For y′=-2(y-2),y(0)=5, the candidate y=2+3e^(−2x) has derivative -6e^(−2x). The right-hand side is also -6e^(−2x), and the initial value is 5. It is defined for all real x. Changing the constant offset can break the differential equation, while changing the exponential coefficient can preserve the equation but alter the initial value

For y=2+3e^(−2x), what is y′ at x=0?

A direction field is local slope data

At each selected point draw a short segment with slope f(x,y). A solution curve should be tangent to the field at every point it passes through. The segment’s drawn length is chosen for clarity; it does not specify a time step or a numerical solution. Numerical stepping comes later in 6D

For the displayed equation y′=2-y, what do the slope segments indicate above y=2?

Is the dashed zero-slope line y=x a solution of y′=x-y?

Form a mixing balance from units

In a perfectly mixed 40 L tank, 4 L/min enters with concentration 1.5 g/L and 4 L/min leaves. Let M be dye mass in grams. Inflow carries 6 g/min; outflow carries 4(M/40)=M/10 g/min. Thus M′=6-M/10, with constant volume because flows match. At M=20 g the rate is 4 g/min; equilibrium mass is 60 g.

A perfectly mixed 40 L tank has a 4 L/min inlet with concentration 1.5 g/L and a 4 L/min outlet. Give dM/dt when the dye mass is 20 g

For the dye-mass model dM/dt=6−M/10, give the equilibrium mass in grams

Signed differences explain warming as well as cooling

For constant ambient 18°C and coefficient 1/5 per minute, T′=-(T-18)/5. At T=13°C this is +1°C/min, so the model warms toward ambient. At T=23°C it is -1°C/min. The same signed rule covers both sides; assume a constant environment and a uniform object temperature

For T′=−(T−18)/5 in °C/min, give the temperature rate at T=13°C

Other rate models need explicit assumptions

A population with continuous per-capita growth r and fixed removal H can be modelled by P′=rP-H. A balance earning continuous interest r with constant deposits d has B′=rB+d; repayments p instead give B′=rB-p. These are modelling assumptions, not discrete annual update rules. State the time unit and stop or revise a model if its physical constraints fail

Find equilibria before discussing long-time behaviour

For autonomous y′=g(y), constant solutions occur at roots of g. In y′=(y-1)(y-3), signs are positive below 1, negative between 1 and 3, and positive above 3. Thus 1 attracts nearby solutions and 3 repels them. A direction field gives qualitative local behaviour; existence for arbitrarily large x still needs checking

Give the upper equilibrium in y′=(y-1)(y-3)

A solution formula can have a smaller domain than the equation

The candidate y=1+2/(x-3) satisfies y′=-(y-1)²/2 wherever x≠3. At x0=1 it gives y0=0 and slope -1/2, so its solution branch uses x<3. The right-hand side itself is defined for every finite y, but this candidate blows up at 3. The separate constant solution y=1 is valid too

For y=1+2/(x−3), give y(1)

For y=1+2/(x−3), give y′(1)

Uniqueness requires hypotheses

A continuous right-hand side with a continuous partial derivative with respect to y near the initial point supplies a standard local uniqueness condition. It does not guarantee that a solution exists for all time. Without suitable hypotheses, curves through the same initial point can differ; field appearance alone is not a proof of uniqueness

Form three contextual equations, ungraded

Form and explain models for: (i) dye mass in a well-mixed tank whose inlet and outlet volume rates differ;(ii) a population with proportional growth and fixed removal;(iii) a loan balance under continuous interest and continuous repayments. Name variables, units, initial conditions and validity limits. No analytic solution is required here

Model: balance the quantity, then state what can invalidate it

For dye mass M(t)g, initial volume V0 L, inlet/outlet qin,qout L/min and inlet concentration c g/L, use V(t)=V0+(qin-qout)t and M′=qin c-qout M/V(t),M(0)=M0, while V(t)>0 and capacity assumptions hold. Perfect mixing and no chemical reaction are assumptions. For P(t) individuals over years, P′=rP-H,P(0)=P0, with r per year,H individuals/year and P≥0. For B(t) dollars over years, B′=rB-p,B(0)=B0, with r per year and p dollars/year; stop at repayment or revise any negative-balance interpretation. Data should test constant-rate assumptions rather than being treated as a proof

Write a full solution verification, ungraded

Verify y=2+3e^(−2x) for y′=-2(y-2),y(0)=5 on its full real domain. Explain exactly what changes if the candidate is 2+4e^(−2x) or 3+3e^(−2x)

Model: compare identities before checking the point

For y′=−2(y−2), y(0)=5, the first candidate y=2+3e^(−2x) is differentiable on all real x; its derivative and substituted right-hand side both equal -6e^(−2x), and y(0)=5. The candidate 2+4e^(−2x) still satisfies the differential identity, but starts at 6. For 3+3e^(−2x), derivative is -6e^(−2x) whereas the right side is -2-6e^(−2x); its residual is 2, so it fails the equation everywhere and starts at 6. A matching plotted point cannot replace this identity check

Construct and interpret a direction field, ungraded

For y′=2-y, construct a small slope grid and sketch solutions through (0,1),(0,2),(0,3). Verify proposed curves, explain why they do not cross, and distinguish the plotted tangent segments from a numerical time-stepping method

Model: use horizontal slope rows and one constant family

At y=0,1,2,3 the slopes are 2,1,0,-1 for every x. The proposed family y=2+C e^(−x) differentiates to -C e^(−x)=2-y. For initial points (0,1), (0,2) and (0,3), the constants are −1, 0 and 1 respectively. The curves stay below, on and above 2, approaching 2 as x increases. The smooth right-hand side gives local uniqueness, consistent with these explicit curves not crossing. A plotted segment is only a tangent indicator; Euler stepping later uses a chosen horizontal increment and computed vertical change

Explain domain and uniqueness limits, ungraded

Verify that y=0 and y=x³ both satisfy y′=3y^(2/3),y(0)=0 for x≥0. Construct a delayed-start solution for a fixed c≥0. Contrast this with the rational candidate’s pole and the local uniqueness criterion

Model: distinguish a failed hypothesis from a singular branch

For x≥0 with initial condition y(0)=0, y=x³ has y′=3x²=3(x³)^(2/3); y=0 also satisfies the equation. Define y=0 for 0≤x≤c and y=(x-c)³ for x≥c. Values and first derivatives match at c, so this is another solution with the same initial value. The y derivative of 3y^(2/3) is unbounded near 0, so the stated sufficient uniqueness condition fails there. In the rational example the right-hand side is smooth, but a particular solution becomes unbounded at a finite x; local uniqueness does not remove that pole or imply global existence

Plan a technology-assisted model check, ungraded

For y′=x-y²,y(0)=1, state a reproducible direction-field and numerical exploration plan, checks and limitations. Explain what you can conclude before finding a closed form and how you would assess a proposed curve

Model: record the equation and compare local behaviour

Use x horizontally and y vertically, plot slopes x-y² on a stated grid and mark (0,1). The initial tangent slope is -1. Zero slopes lie on x=y² with x≥0, but those curves are not automatically solutions. Use a documented numerical solver and tolerances, repeat with finer settings and compare against the local slopes; 6D supplies an explicit Euler method. Record the explored interval and stop if values become unbounded or leave the model domain. For any proposed expression compute y′-(x-y²) symbolically over a connected domain and check y(0)=1. Numerical agreement at finitely many points is evidence, not an identity proof

Assessment boundary

New practice grades bounded substitutions, residual coefficients, balance coefficients, local signs and domain choices. Complete contextual formulation, verification arguments, direction-field construction and independent technology execution remain ungraded. The companion lesson practises initial values and derivatives of supplied exponential functions; it does not ask you to solve new differential equations