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ODE, PDE & Applied Mathematics

Numerical Analysis

18 previous-year questions →every option and the correct answer, freePractice this topic →one question at a time, no clock, check as you go

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1. Root finding: bisection, Newton–Raphson, fixed point, order of convergence

Read this as a chapter →

Lec-26 Solution of Non-linear Equations

NPTEL · Elementary Numerical Analysis

Bisection, fixed point and Newton in one place.

Lec-27 Quadratic Convergence of Newton's Method

NPTEL · Elementary Numerical Analysis

Why Newton is quadratic — and only at simple roots.

Root finding and orders of convergence free

In plain words

Bisection is slow and cannot fail. Newton roughly doubles the number of correct digits each step, but only at a simple root, and only when you start close enough.

The comparison table

Method Order Needs Guaranteed?
Bisection 1 (linear, factor ½) sign change ✓ always
Regula falsi 1 sign change ✓
Fixed point xnx_{n}₊1=g(xn)_{1} = g(x_{n}) 1 if 0 < |g′(r)| < 1; 2 if g′(r) = 0 |g′| < 1 near r local
Secant (1+5)/2≈1.618(1+\sqrt{5})/2 \approx 1.618 two starting points local
Newton xnx_{n}₊1=xn−f/f′_{1} = x_{n} - f/f' 2 at a simple root, 1 at a multiple root f′ ≠ 0 local only
Modified Newton x − m f/f′ 2 at a root of known multiplicity m local

Newton's caveats

  • Quadratic convergence is local: a poor start can diverge (f = x1/3)x^{1/3}), cycle (f=x3−2x+2(f = x^{3} - 2x + 2 from x0=0)x_{0} = 0), or run away.
  • At a root of multiplicity m ≥ 2 the convergence degrades to linear with factor (m−1)/m.
  • Error recursion: en+1e_{n+1} ≈(f′′/2f′)en2\approx (f''/2f') e_{n}^{2}.

Fixed point iteration

Converges locally iff |g′(r)| < 1; the error satisfies en+1e_{n+1} ≈g′(r)en\approx g'(r)e_{n}. Banach's contraction principle gives a global statement on an interval where |g′| ≤ k < 1 and g maps the interval into itself.

Bisection count

To reach tolerance ε\varepsilon from an interval of length L:n≥log2(L/ε)L: n \ge log_{2}(L/\varepsilon) steps. Bisection never fails but never accelerates.

Key takeaways

  • Newton is quadratic only at simple roots, and only locally.
  • Secant beats bisection in order but loses the guarantee.
  • Fixed point: the derivative at the root is the convergence factor.

Newton's method: racing — and cyclinginteractive

This visual is part of Trap Analysis

Slide the starting point: quadratic convergence one moment, an eternal 2-cycle the next. Local convergence, felt.

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Exam focus: Order of convergence is the whole topic: bisection 1, secant φ ≈ 1.618, Newton 2 (but only 1 at a multiple root), fixed point 1 unless g′(r) = 0.

Practice

Part C2025 Decembershow ▾
Let {xnx_{n}} be a convergent iterative sequence generated by Newton-Raphson method for solving the equation sin x−1=0 such that xn→π/2x_{n}\to\pi/2 as n→∞n\to\infty. For n∈Nn\in\mathbb{N}, let en=xn−π/2e_{n}=x_{n}-\pi/2. Let p>0 be such that lim⁡(n→∞)\lim(n\to\infty)|ene_{n}₊1_{1}|/|ene_{n}|^p exists and is non-zero. Then which of the following statements are true?
  1. A.p = 1✓ Correct
  2. B.lim⁡(n→∞)\lim(n\to\infty) |ene_{n}₊1_{1}|/|ene_{n}|^p = 1/2✓ Correct
  3. C.p = 2
  4. D.lim⁡(n→∞)\lim(n\to\infty) |ene_{n}₊1_{1}|/|ene_{n}|^p = 1

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾
Let S be the set of all 2×2 matrices A such that the iterative sequence generated by the Gauss-Seidel method converges for every initial guess, when employed to solve the system of equations A(x1,x2)T=(1,2)TA(x_{1},x_{2})^{T}=(1,2)^{T}. Then which of the following statements are true?
  1. A.(4,1;2,3) ∈ S✓ Correct
  2. B.(1,2;3,4) ∈ S
  3. C.(1,5;1,10) ∈ S✓ Correct
  4. D.(−5,2;1,−4) ∈ S✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾
Let f:R→Rf : \mathbb{R} \to \mathbb{R} be such that sup(x≠y) |f(x) − f(y)|/|x − y| = L, where 1<L<∞1 < L < \infty. Let h:R→Rh : \mathbb{R} \to \mathbb{R} be a differentiable function satisfying |h′(x)| ≤ 3/4 for all x∈Rx \in \mathbb{R}. For α>0\alpha > 0, define g(x)=αf(x)+h(x)g(x) = \alpha{}f(x) + h(x) for x∈Rx \in \mathbb{R}. Consider the sequence {xkx_{k}}(k≥0) defined by x(k+1)=g(xk),k=0,1,…x_(k+1) = g(x_{k}), k = 0, 1, \dots, where x0∈Rx_{0} \in \mathbb{R}. The sequence {xkx_{k}}(k≥0) converges to the solution of the equation x = g(x) if
  1. A.α<2/(3L)\alpha < 2/(3L)
  2. B.α<3/(2L)\alpha < 3/(2L)
  3. C.α<4L\alpha < 4L
  4. D.α<1/(4L)\alpha < 1/(4L)✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾
Consider the following statements: S1S_{1}: There exists an ε>0\varepsilon > 0 such that ∀x0∈(2−ε,2+ε)\forall{}x_{0} \in (2 - \varepsilon, 2 + \varepsilon), the iterative sequence defined by x(n+1)=6/(5−xn),n=0,1,2,3,…x(n+1) = 6/(5 - x_{n}), n = 0, 1, 2, 3, \dots, converges to 2.S22. S_{2}: There exists an ε>0\varepsilon > 0 such that ∀x0∈(3−ε,3+ε)\forall{}x_{0} \in (3 - \varepsilon, 3 + \varepsilon), the iterative sequence defined by x(n+1)=(xn2+6)/5,n=0,1,2,3,…x(n+1) = (x_{n}^{2} + 6)/5, n = 0, 1, 2, 3, \dots, converges to 3.S33. S_{3}: There exists an ε>0\varepsilon > 0 such that ∀x0∈(3−ε,3+ε)\forall{}x_{0} \in (3 - \varepsilon, 3 + \varepsilon), the iterative sequence defined by x(n+1)=(5xn−6)/xn,n=0,1,2,3,…x(n+1) = (5x_{n} - 6)/x_{n}, n = 0, 1, 2, 3, \dots, converges to 3. Then which of the following statements are true?
  1. A.S1S_{1} and S2S_{2} are true but NOT S3S_{3}
  2. B.S2S_{2} and S3S_{3} are true but NOT S1S_{1}
  3. C.S1S_{1} and S3S_{3} are true but NOT S2S_{2}✓ Correct
  4. D.Each of S1,S2S_{1}, S_{2}, and S3S_{3} is true

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾
Let S denote the set of all 2 × 2 matrices A such that the iterative sequence generated by the Gauss-Seidel method applied to the system of linear equations A(x1,x2)T=(2,3)TA(x_{1}, x_{2})^{T} = (2, 3)^{T} converges for every initial guess. Then which of the following statements are true?
  1. A.[[5, 8], [1, 2]] ∈ S✓ Correct
  2. B.[[3, 2], [1, 2]] ∈ S✓ Correct
  3. C.[[−3, 1], [2, 3]] ∈ S✓ Correct
  4. D.[[2, 2], [4, 3]] ∈ S

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾
Consider the fixed-point iteration xnx_{n}₊1=g(xn)_{1} = g(x_{n}) near a fixed point x* = g(x*), with g continuously differentiable. Which of the following are true?
  1. A.If |g′(x*)| < 1, the iteration converges from every start sufficiently close to x*.✓ Correct
  2. B.If |g′(x*)| > 1, the iteration cannot converge to x* from any starting point whatsoever.
  3. C.If g′(x*) = 0 and g″ is continuous, the local convergence is at least quadratic.✓ Correct
  4. D.Newton's method for a simple root of f is a fixed-point iteration whose g′ vanishes at the root.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾
Which of the following are true about Newton–Raphson for a simple root?
  1. A.Convergence is quadratic near the root✓ Correct
  2. B.It always converges from any starting point
  3. C.For a double root convergence degrades to linear✓ Correct
  4. D.It requires f′ at each iterate✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾
Which of the following statements about the Newton–Raphson method for solving f(x) = 0 are true?
  1. A.Near a simple root, with a sufficiently good initial guess, the convergence is quadratic.✓ Correct
  2. B.At a root of multiplicity 2, the convergence is only linear.✓ Correct
  3. C.The iteration converges for every choice of initial guess.
  4. D.Like bisection, it requires an interval on which f changes sign.

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾
Applying Newton's method to f(x)=x2−2f(x) = x^{2} - 2 with x0=1x_{0} = 1, the first iterate x1x_{1} is
  1. A.1.25
  2. B.1.4
  3. C.1.5✓ Correct
  4. D.2

The worked solution and the per-option analysis are in Trap Analysis.

2. Interpolation and numerical integration with error terms

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Lec-03 Interpolating Polynomials

NPTEL · Elementary Numerical Analysis

Lec-05 Error in the Interpolating polynomial

NPTEL · Elementary Numerical Analysis

The f⁽ⁿ⁺¹⁾/(n+1)! error term.

Lec-10 Numerical Integration: Basic Rules

NPTEL · Elementary Numerical Analysis

Trapezoidal and Simpson, with degrees of precision.

Lec-12 Gauss 2-point Rule: Construction

NPTEL · Elementary Numerical Analysis

Why n Gauss points achieve degree 2n − 1.

Interpolation and numerical integration free

In plain words

Exactly one polynomial passes through your points. With equally spaced points, adding more can make the fit worse, so the fix is to move the points rather than add them.

Interpolation

Through n + 1 distinct nodes there is a unique polynomial of degree ≤ n.

  • Lagrange form: p(x)=∑yi∏j≠i(xp(x) = \sum y_{i} \prod_{j\ne{}i}(x −xj)/(xi−xj)- x_{j})/(x_{i} - x_{j}).
  • Newton divided differences: p(x)=f[x0]+f[x0,x1](x−x0)+⋯p(x) = f[x_{0}] + f[x_{0},x_{1}](x - x_{0}) + \cdots; the leading coefficient is f[x0,…,xn]f[x_{0}, \dots, x_{n}].
  • Error: f(x) − p(x) = f⁽n^{n}⁺1^{1}⁾(ξ)/(n+1)!⋅∏(x−xi)(\xi)/(n+1)! \cdot \prod(x - x_{i}). −f[x0,…,xn]=f- f[x_{0}, \dots, x_{n}] = f⁽n^{n}⁾(ξ)/n(\xi)/n! — so divided differences of order > deg f vanish.
  • Runge's phenomenon: equally spaced nodes for 1/(1+25x2)1/(1 + 25x^{2}) diverge as n grows; Chebyshev nodes fix it.

Quadrature

Rule Error term Degree of precision
Trapezoidal −(b−a)3f′′(ξ)/12-(b-a)^{3}f''(\xi)/12 1
Simpson's 1/3 −(b−a)5f-(b-a)^{5}f⁽4^{4}⁾(ξ)/2880(\xi)/2880 3
Simpson's 3/8 O(h5)O(h^{5}) 3
Midpoint (b−a)3f′′(ξ)/24(b-a)^{3}f''(\xi)/24 1
n-point Gauss–Legendre involves f⁽2n^{2n}⁾ 2n − 1

Composite rules: trapezoidal error O(h2)O(h^{2}), Simpson O(h4)O(h^{4}).

Simpson is exact for cubics despite being built from a quadratic fit — the standard "surprising" exam fact. Two-point Gauss (nodes ±1/3)\pm1/\sqrt{3}) is exact for cubics too.

Key takeaways

  • n + 1 nodes ⇒ unique interpolant of degree ≤ n; error has f⁽n^{n}⁺1^{1}⁾/(n+1)!.
  • Simpson has degree of precision 3, not 2.
  • Gauss with n nodes achieves 2n − 1 — the best possible.

Degree of exactness, and why more nodes can be worseinteractive

This visual is part of Trap Analysis

Trapezium versus Simpson, and Runge's phenomenon in one example.

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Exam focus: Degree of precision is the key number: trapezoidal 1, Simpson 3, n-point Gauss 2n−1. Interpolation error carries f⁽ⁿ⁺¹⁾/(n+1)! times the node product.

Practice

Part C2025 Juneshow ▾
If α,β∈R\alpha, \beta \in \mathbb{R} are such that the equation ∫03f(x)\int_{0}^{3} f(x) dx =(3/2)[f(α)+f(α+β)]= (3/2)[f(\alpha) + f(\alpha + \beta)] holds for all polynomials f(x) of degree less than or equal to 2, then which of the following statements are true?
  1. A.(α,β)=((3−3)/2,3)(\alpha, \beta) = ((3 - \sqrt{3})/2, \sqrt{3}) or (α,β)=((3+3)/2,−3)(\alpha, \beta) = ((3 + \sqrt{3})/2, -\sqrt{3})✓ Correct
  2. B.(α,β)=((3−2)/2,2)(\alpha, \beta) = ((3 - \sqrt{2})/2, \sqrt{2}) or (α,β)=((3+2)/2,−2)(\alpha, \beta) = ((3 + \sqrt{2})/2, -\sqrt{2})
  3. C.(α,β)=((3−5)/2,5)(\alpha, \beta) = ((3 - \sqrt{5})/2, \sqrt{5}) or (α,β)=((3+5)/2,−5)(\alpha, \beta) = ((3 + \sqrt{5})/2, -\sqrt{5})
  4. D.(α,β)=((3−7)/2,7)(\alpha, \beta) = ((3 - \sqrt{7})/2, \sqrt{7}) or (α,β)=((3+7)/2,−7)(\alpha, \beta) = ((3 + \sqrt{7})/2, -\sqrt{7})

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾
If the function s:[0,4]→Rs : [0, 4] \to \mathbb{R} defined by s(x)=a(x−2)2+b(x−1)2s(x) = a(x - 2)^{2} + b(x - 1)^{2} for 0≤x≤1,s(x)=(x−2)20 \le x \le 1, s(x) = (x - 2)^{2} for 1 < x ≤ 3, and s(x)=2c(x−2)2+(x−3)3s(x) = 2c(x - 2)^{2} + (x - 3)^{3} for 3 < x ≤ 4, is a cubic spline, then the value of 2a + b + 2c is
  1. A.2
  2. B.3✓ Correct
  3. C.4
  4. D.5

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Decembershow ▾
If p(x) is the interpolating polynomial for the data x = −2,−1,0,1,2 with y = −1,3,1,−1,3, then the value of p(1/2) is
  1. A.−3/8✓ Correct
  2. B.−5/8
  3. C.5/8
  4. D.3/8

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾
If λ∈R\lambda \in \mathbb{R} and p∈Rp \in \mathbb{R} are such that the quadrature formula ∫(x0)(x0+h)f(x)\int_(x_{0})^(x_{0}+h) f(x)dx ≈λh(f(x0)+f(x0+h))+\approx \lambda{}h(f(x_{0}) + f(x_{0} + h)) + ph3(f′′(x0)+f′′(x0+h))^{3}(f''(x_{0}) + f''(x_{0} + h)) is exact for all polynomials of degree as high as possible, then
  1. A.2λ+24p=02\lambda + 24p = 0✓ Correct
  2. B.7λ−12p=47\lambda - 12p = 4✓ Correct
  3. C.2λ+24p=−32\lambda + 24p = -3
  4. D.7λ−12p=117\lambda - 12p = 11

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾
Let f(x) be the polynomial of degree at most 2 that interpolates the data (−1, 2), (0, 1), and (1, 2). If g(x) is a polynomial of degree at most 3 such that f(x) + g(x) interpolates the data (−1, 2), (0, 1), (1, 2), and (2, 17), then
  1. A.f(5) + g(3) = 50
  2. B.2f(5) − g(3) = 4✓ Correct
  3. C.f(1) + g(3) = 50✓ Correct
  4. D.f(5) + g(3) = 74✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾
Let g(x) be the polynomial of degree at most 4 that interpolates the data (x, y) = (−1, −30), (0, 1), (2, c), (3, 10), (6, 19). If g(4) = 5, then which of the following statements are true?
  1. A.c = 13
  2. B.g(5) = 6✓ Correct
  3. C.g(1) = 14✓ Correct
  4. D.c = 15✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾
Consider the quadrature formula ∫\int₋11_{1}^{1} |x|f(x)dx ≈ (1/2)(f(−1) + f(1)). Then the degree of precision (also known as order of exactness) of the quadrature formula is
  1. A.0
  2. B.1✓ Correct
  3. C.2
  4. D.3

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾
The coefficient of x3x^{3} in the interpolating polynomial for the data x = 0, 1, 2, 3, 4 with y = 1, 2, 1, 3, 5 is
  1. A.−1/3
  2. B.−1/2
  3. C.5/6
  4. D.17/6✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾
Which of the following statements about polynomial interpolation and quadrature are true?
  1. A.The interpolating polynomial of degree ≤ n through n + 1 points with distinct nodes is unique.✓ Correct
  2. B.Simpson's rule integrates every cubic polynomial exactly.✓ Correct
  3. C.The single-interval trapezoidal rule has error −(b−a)3f′′(ξ)/12-(b - a)^{3}f''(\xi)/12 for some ξ∈(a,b)\xi \in (a, b).✓ Correct
  4. D.Increasing the number of equally spaced interpolation nodes always reduces the maximum interpolation error.

The worked solution and the per-option analysis are in Trap Analysis.

Part Bshow ▾
The degree of exactness of the (single-interval) trapezoidal rule is
  1. A.0
  2. B.1✓ Correct
  3. C.2
  4. D.3

The worked solution and the per-option analysis are in Trap Analysis.

3. Numerical ODE: Euler, Runge–Kutta

Read this as a chapter →

Lec-02 Single - Step Methods for IVPs

NPTEL · Numerical Methods of ODE and PDE

Euler and the local-vs-global order distinction.

Lec-04 Runge - Kutta Methods for IVPs

NPTEL · Numerical Methods of ODE and PDE

Where the RK2 order conditions come from.

Lec-06 Error - Stability - Convergence of Single Step Methods

NPTEL · Numerical Methods of ODE and PDE

Stability regions — why explicit Euler needs a small step.

Numerical methods for ODE free

In plain words

Step along the slope and repeat. Accuracy and stability are different things: too large a step makes an explicit method grow where the true solution decays.

The methods

Method Formula Global order
Forward Euler yny_{n}₊1=yn+hf(xn,yn)_{1} = y_{n} + h f(x_{n}, y_{n}) 1
Backward Euler yny_{n}₊1=yn+hf(xn_{1} = y_{n} + h f(x_{n}₊1,yn_{1}, y_{n}₊1)_{1}) 1 (implicit, A-stable)
Trapezoidal / Heun yny_{n}₊1=yn+(h/2)(f(xn,yn)+f(xn_{1} = y_{n} + (h/2)(f(x_{n},y_{n}) + f(x_{n}₊1_{1}, ŷ)) 2
RK4 classical four-stage 4

Local truncation error is one order higher than the global error.

RK2 order conditions

y1=y0+y_{1} = y_{0} + Pk1+_{1} + Qk2_{2} with k1=hf(x0,y0),k2=hf(x0+αh,y0+βk1)k_{1} = h f(x_{0},y_{0}), k_{2} = h f(x_{0} + \alpha{}h, y_{0} + \beta{}k_{1}) is second order iff

P+Q=1,Qα=P + Q = 1, Q\alpha = ½, Qβ=Q\beta = ½ (so α=β\alpha = \beta and Q=1/(2α))Q = 1/(2\alpha)).

α=1\alpha = 1 gives Heun (P = Q = ½);α=); \alpha = ½ gives the midpoint method (P=0,Q=1);α=2(P = 0, Q = 1); \alpha = 2 gives P = ¾, Q = ¼.

Stability

Applied to y′=λy(Reλ<0)y' = \lambda{}y (Re \lambda < 0) with z=hλz = h\lambda:

  • Forward Euler is stable for |1 + z| < 1, i.e. real λ\lambda needs h < 2/|λ\lambda|.
  • Backward Euler and the trapezoidal rule are A-stable (stable for all h).
  • Stiff problems (widely separated eigenvalues) force implicit methods.

Worked Euler

y′=3x+2y+1,y(1)=1,h=0.05:y1=1+0.056≈1.1225,y2≈1.1225+0.055.395≈1.25y' = \sqrt{3x + 2y + 1}, y(1) = 1, h = 0.05: y_{1} = 1 + 0.05\sqrt{6} \approx 1.1225, y_{2} \approx 1.1225 + 0.05\sqrt{5.395} \approx 1.25.

Key takeaways

  • Global order = local order − 1.
  • RK2 conditions: P+Q=1,Qα=Qβ=P + Q = 1, Q\alpha = Q\beta = ½.
  • Explicit methods have bounded stability regions; implicit ones can be A-stable.

Order of a method, and the step size stability forces on youinteractive

This visual is part of Trap Analysis

Why explicit Euler needs h < 2/|λ| however accurate you wanted to be.

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Exam focus: Know the local vs global order (Euler: local O(h²), global O(h)), the RK2 order conditions, and the stability interval for Euler applied to y′ = λy.

Practice

Part C2025 Juneshow ▾
Consider the initial value problem (IVP) y′ + y = 0, y(0) = 1. Let (yn)(y_{n}) be the iterates of forward Euler method, applied to the IVP, with step size h where 0 < h < 1. Then which of the following statements are true?
  1. A.The sequence (yn)(y_{n}) does NOT converge
  2. B.yn→0y_{n} \to 0 as n→∞n \to \infty✓ Correct
  3. C.0≤yn≤10 \le y_{n} \le 1 for n = 0, 1, 2, …✓ Correct
  4. D.|y(nh)−yn) - y_{n}| → 0 as n→∞n \to \infty✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾
Let f:[0,1]×R→Rf : [0, 1] \times \mathbb{R} \to \mathbb{R} be a function such that f and its partial derivatives of orders less than or equal to 3 are continuous and bounded. Let y:[0,1]→Ry : [0, 1] \to \mathbb{R} be the solution of dy/dt =f(t,y(t)),0≤t≤1,y(0)=y0= f(t, y(t)), 0 \le t \le 1, y(0) = y_{0}, where y0∈Ry_{0} \in \mathbb{R}. For h > 0, denote t_j = jh. Let Y_j be an approximation of y(t_j), defined by Y_(j+1) = Y_j + ahf(t_j, Y_j) + bhf(t_(j+1), Y_j + chf(tj,Yj)),0<(j+1)h<1,Y0=y0(t_j, Y_j)), 0 < (j+1)h < 1, Y_{0} = y_{0}, where a,b,c∈Ra, b, c \in \mathbb{R}. If there exists an M > 0 such that |y(t_(j+1)) − y(t_j) − ahf(t_j, y(t_j)) − bhf(t_(j+1), y(t_j) + chf(t_j, y(t_j)))| ≤ Mh3^{3} for every h > 0, and every j with 0 ≤ (j+1)h < 1, then
  1. A.a + b = 1✓ Correct
  2. B.a + b = c✓ Correct
  3. C.a + b + c = 0
  4. D.a2+b2=c2a^{2} + b^{2} = c^{2}

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾
If the value of the approximate solution of the initial value problem y′(x)=x(y(x)+1),x∈R,y(0)=βy'(x) = x(y(x) + 1), x \in \mathbb{R}, y(0) = \beta at x = 0.2 using the forward Euler method with step size 0.1 is 1.02, then the value of β\beta is
  1. A.0
  2. B.−1
  3. C.2
  4. D.1✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾
For the IVP dy/dx =f(x,y),y(x0)=y0= f(x, y), y(x_{0}) = y_{0}, with step size h, let the first iterate of a second-order scheme be y1=y0+y_{1} = y_{0} + Pk1+_{1} + Qk2_{2}, where k1=hf(x0,y0),k2=hf(x0+α0h,y0+β0k1)k_{1} = h f(x_{0}, y_{0}), k_{2} = h f(x_{0} + \alpha_{0}h, y_{0} + \beta_{0}k_{1}) and P,Q,α0,β0∈RP, Q, \alpha_{0}, \beta_{0} \in \mathbb{R}. Which of the following statements are correct?
  1. A.If α0=2\alpha_{0} = 2, then β0=2,P=3/4,Q=1/4\beta_{0} = 2, P = 3/4, Q = 1/4✓ Correct
  2. B.If β0=3\beta_{0} = 3, then α0=3,P=5/6,Q=1/6\alpha_{0} = 3, P = 5/6, Q = 1/6✓ Correct
  3. C.If α0=2\alpha_{0} = 2, then β0=2,P=1/4,Q=3/4\beta_{0} = 2, P = 1/4, Q = 3/4
  4. D.If β0=3\beta_{0} = 3, then α0=3,P=1/6,Q=5/6\alpha_{0} = 3, P = 1/6, Q = 5/6

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾
Using Euler's method with step size 0.05, the approximate value of the solution of dy/dx =3x+2y+1,y(1)=1= \sqrt{3x + 2y + 1}, y(1) = 1, at x = 1.1 (rounded to two decimal places) is
  1. A.1.50
  2. B.1.65
  3. C.1.25✓ Correct
  4. D.1.15

Solution

y1=1+0.056≈1.1225;y2=1.1225+0.053.15+2.245+1≈1.1225+0.05⋅2.529≈1.249≈1.25y_{1} = 1 + 0.05\sqrt{6} \approx 1.1225; y_{2} = 1.1225 + 0.05\sqrt{3.15 + 2.245 + 1} \approx 1.1225 + 0.05\cdot2.529 \approx 1.249 \approx 1.25.

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Part Cshow ▾
Consider solving y′ = f(t, y) numerically with step size h. Which of the following are true?
  1. A.Explicit Euler has local truncation error O(h2)O(h^{2}) and global error O(h).✓ Correct
  2. B.Classical fourth-order Runge–Kutta has global error O(h4)O(h^{4}).✓ Correct
  3. C.Halving h in explicit Euler roughly halves the global error.✓ Correct
  4. D.For y′=λyy' = \lambda{}y with real λ<0\lambda < 0, explicit Euler is stable for every step size h > 0.

Solution

(1)/(2) are the standard orders — the global order sits one below the local order because ~1/h steps accumulate.

(3) restates global order 1.

(4) fails: explicit Euler gives yny_{n}₊1=(1+hλ)yn_{1} = (1 + h\lambda)y_{n}, stable only when |1+hλ1 + h\lambda| < 1, i.e. h < 2/|λ\lambda|.

That restriction is what stiffness means; implicit Euler removes it.

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