Approximate solutions with explicit Euler steps

Start from the rate and the initial point

Senior 6A supplies f(x,y) and direction fields. Unit 1 3D supplies sequential assignment and finite loops. Euler combines them: follow a tangent for a chosen horizontal step, then recompute the slope at the new approximate point. The initial value is known; later values are generally approximate

Derive the first-order step

For a differentiable solution, y(x+h) is approximately y(x)+h y′(x). Replace y′ by f(x,y) and use the current numerical value: update the horizontal coordinate and use the current numerical point to update the vertical coordinate. Both old coordinates enter the slope. The approximation sign belongs to comparison with the exact solution, not to the arithmetic of the recurrence

Work the first tangent step

For y′=x+y, y(0)=1 and h=1/2, the first slope is 0+1=1. The vertical increment is (1/2)×1=1/2, so (x1,y1)=(1/2,3/2). A slope is a rate of change, not the next y value. Units of h times the derivative must match units of y.

For y′=x+y, y(0)=1 and h=1/2, give the first Euler y value

Count intervals and stored points

To cover [x0,T] in N equal forward steps, divide the interval length by N to get the step size, with N a positive integer. Store indices 0 through N: there are N updates and N+1 points. If a chosen h does not divide the interval, explicitly choose a shortened final step or a new uniform h; do not silently overshoot or compare different endpoints

For x0=0,T=1,N=4, give h.

How many points are stored for four Euler updates, including the initial point?

Recompute the slope at the approximate point

Continue y′=x+y from (1/2,3/2). The second slope is 2, so y2=3/2+(1/2)(2)=5/2 at x2=1. Reusing the initial slope would give a different result. The exact solution 2e^x-x-1 has y(1)=2e-2, so the Euler endpoint is an approximation

What do the straight segments in this graph represent?

Linear growth becomes a different multiplier

For y′=ky, Euler gives y[n+1]=(1+kh)y[n]. The exact continuous solution advances by e^(kh). With y0=2,k=1,h=1/2, three Euler values are 3,9/2,27/4. Do not replace 1+kh by the exact multiplier when the task specifically asks for Euler

For y′=y, y(0)=2 and h=1/2, give the third Euler y value

Nonautonomous rules depend on both coordinates

For y′=x-y, y(0)=1 and h=1/2, the first slope is -1 and y1=1/2. At (1/2,1/2), the second slope is zero and y2=1/2+(1/2)×0=1/2. At (1,1/2), the third slope is 1/2 and y3=3/4. Use the stated h=1/2 in every increment

Give the third Euler y value for y′=x-y, y(0)=1, h=1/2.

Step a nonlinear logistic rate

For P′=(1/2)P(1-P/120), P0=30 and h=1 day, the first rate is 45/4, giving P1=165/4. Recompute the rate at 165/4: it is 3465/256, so P2=14025/256. These are approximate populations in thousands of cells. The exact reference at t=2 is 120/(1+3e^(-1)); equality is not expected

For P′=(1/2)P(1−P/120), P₀=30 and h=1 day, give the first Euler population value, in thousands of cells

Compare at the same endpoint

For y′=2x,y(0)=1 over [0,1], h=1/2 gives slopes 0 and 1 and endpoint 3/2. Four steps with h=1/4 use slopes 0,1/2,1,3/2 and give 7/4. Integration gives exact y=1+x² and endpoint 2. Signed errors, numerical minus exact, are -1/2 and -1/4.

The coarse and fine absolute errors are 1/2 and 1/4 respectively. Give their ratio

Distinguish one-step and accumulated errors

Taylor expansion gives a one-step value defect of order h² when starting from an exact point and the second derivative is bounded. On a fixed finite interval, appropriate smoothness and stability control propagation and yield Euler global error O(h). The normalised local truncation error is sometimes defined after division by h; state the convention. Halving h usually reduces truncation error, but no universal monotone-error claim follows for every problem and machine calculation

Assignment order can change the numerical method

For y′=x+y at (0,1), h=1/2, correct Euler gives 3/2. Faulty code x←x+h; y←y+h(x+y) uses the new x=1/2, so it gives 7/4. Compute the old-point slope before replacing x or y. A single test where f does not depend on x would miss this bug

A faulty update gives 7/4 and the correct Euler update gives 3/2. Give faulty minus correct

Numerical instability can contradict the exact model

For y′=-2y, y(0)=4, exact y=4e^(-2x) stays positive and decays. With h=3/4, Euler factor 1-2h=-1/2 produces -2,1,-1/2,...: it tends to zero but alternates sign. With h=1 the factor is -1 and magnitude persists; with h=5/4 it is -3/2 and magnitude grows. Numerical instability does not imply that the exact solution grows

Give the Euler amplification factor for y′=-2y with h=3/4.

Use an exact solution to test accumulated error

Let φ(x)=x²+1 and y′=2x+(y-φ(x)), y(0)=1. The exact solution is φ because its correction term vanishes. Euler with h=1/2 gives y1=1, then y2=11/8 at x=1. Exact y(1)=2, so the signed endpoint error is -5/8. After leaving the exact curve, Euler must include its nonzero correction term

At x=1, the Euler value is 11/8 and the exact value is 2. Give Euler minus exact

Respect domains and model meaning

A finite number from a recurrence does not allow crossing a pole or treating a negative population as physical. Record the valid time and state domain, stop when it is left, and reconsider the step or model. A smaller step comparison gives numerical evidence on the tested interval; it does not verify real-world assumptions

Technology: an explicit forward implementation

The following complete JavaScript function returns [x,y] rows including the initial point. It accepts positive h and at most 100000 updates, checks finite arithmetic and reads the old-point slope before assignment. The cap is a practical bound for this example, not a convergence theorem. A domain-specific stopping rule must be added when the model needs one

Technology: runnable source

function euler(f, x0, y0, h, N) { if (typeof f !== 'function' || ![x0,y0,h].every(Number.isFinite) || h <= 0 || !Number.isInteger(N) || N < 0 || N > 100000) throw new Error('Invalid forward grid'); let x = x0, y = y0; const rows = [[x,y]]; for (let n = 0; n < N; n++) { const slope = f(x,y); const nextY = y + h*slope; const nextX = x0 + (n+1)*h; if (![slope,nextX,nextY].every(Number.isFinite)) throw new Error('Non-finite step'); x = nextX; y = nextY; rows.push([x,y]); } return rows; }

Technology: reproduce the reference run

Run euler((x,y)=>x+y,0,1,0.5,2). The rows must be [[0,1],[0.5,1.5],[1,2.5]]. Run euler((x,y)=>x+y,0,1,0.25,4) and check the endpoint [1,369/128]. Compare with exact 2*Math.exp(1)-2. Record both h and N so both runs end at x=1. JavaScript uses floating arithmetic; retain full internal precision and round only the displayed output

Derive and explain the step, ungraded

Starting with y′=f(x,y), derive explicit Euler from a tangent approximation. Explain the old-point convention, the number of stored points and the distinction between a one-step value defect and accumulated error. State sufficient conditions before claiming an error order

Model: a tangent step and controlled propagation

Taylor expansion at an exact point gives y(x+h)=y(x)+h f(x,y(x))+(h²/2)y″(ξ) when the required derivatives exist. Euler drops the final term, so the one-step value defect is O(h²) if y″ is bounded. If f is Lipschitz in y on the relevant region, the difference recurrence is bounded by (1+Lh) times the previous error plus a constant times h²; over a fixed interval this gives an O(h) global bound. These assumptions must hold along the relevant solution and approximations. N updates store N+1 points including index zero

Implement, refine and document a numerical solution, ungraded

Implement Euler for y′=x+y², y(0)=1 on [0,1/5] with h=1/10 and h=1/20. Record every row, endpoint differences, update order and stopping conditions. Test your implementation first on a constant derivative and a known exact solution. Explain why agreement alone is not proof of accuracy

Model: test independently before comparing refinements

Use the supplied function with f=(x,y)=>x+y*y. The h=1/10 run has y values 1,11/10,1231/1000. Its second slope is 1/10+(11/10)²=131/100. The h=1/20 run ends at approximately 1.2503765426223064, while the coarse result is 1.231. Check both endpoints equal 1/5, compare additional refinements and stop if values leave a chosen finite domain. For f=3, y0=2, h=1/4,N=4, the endpoint must be 5; for y′=2x, compare against 1+x². The tests check implementation on their inputs; they do not provide a rigorous error bound for x+y²

Analyse decay stability and physical plausibility, ungraded

For y′=-2y, y(0)=4, classify Euler with h=1/4, 3/4, 1 and 5/4. Give a convergence condition for a general positive h and identify the stronger condition ensuring nonnegative approximations. Compare with the exact model

Model: inspect the recurrence multiplier

The multiplier is q=1-2h. For h=1/4 it is 1/2: positive monotone decay. For h=3/4 it is -1/2: alternating decay. For h=1 it is -1: persistent alternation. For h=5/4 it is -3/2: growing alternation. Convergence to zero requires |q|<1, hence 0<h<1. For this positive start, nonnegative values at every step require q≥0, hence 0<h≤1/2; h=1/2 reaches zero numerically in one step even though exact 4e^(-2x) stays positive. These conditions concern this linear problem, not every differential equation

Investigate a singular equation, ungraded

Consider y′=y², y(0)=1 and the proposed solution y=1/(1−x). Verify this solution and find its connected domain, then explain why stepping from x=0 to x=2 with a finite Euler table cannot justify a continuous solution through x=1.

Model: the pole limits the comparison

The solution is y=1/(1-x) on the maximal interval x<1 containing zero. Differentiating gives y′=1/(1-x)²=y². With h=1, Euler produces y1=2 and y2=6, both finite, although the exact branch has a pole at x=1. Finite arithmetic has not removed that singularity. Restrict comparison to an interval with endpoint below 1, examine the growth and refine the step; do not connect the table across the pole as if it were a classical solution

Assessment boundary

Practice grades bounded updates, trace values, exact-reference comparisons and supplied stability interpretations. Independent code, general convergence arguments, domain justification and physical model validation remain ungraded