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UPSC 2026 Maths Optional Paper 2 Q8b — Step-by-Step Solution

15 marks · Section B

Runge-Kutta methods (RK2/RK4) · Numerical Analysis · asked 5× in 14 yrs · Read the full method →

Question

Consider the initial value problem:

(i) Apply the classical fourth order Runge-Kutta method to approximate the solution $y(2)$.

(ii) Use Euler’s method to find $y(2)$ and $y(3)$.

Technique

Straight application of two standard one-step formulae, but with a step size $h = 1$ that is deliberately enormous — that is the whole point of the question, and the marks are in the arithmetic being exact and completely shown. Note that $f(x,y) = 1 + y/x$ depends on both arguments, so the Runge–Kutta stages must advance $x$ to $1.5$ and $2$ as well as $y$; ignoring the $x$-shift is the standard way to lose this question. The equation is also linear ($y' - y/x = 1$), so the exact solution is available in one line and the comparison table it makes possible is what lifts the answer above a bare numerical grind.

Solution

Step 0 — The problem is well posed on the interval we use

Here

The right-hand side has a singular point at $x = 0$, where $f$ is undefined. But every abscissa used below lies in $[1,3]$: the Euler nodes are $x = 1, 2, 3$ and the Runge–Kutta stages sample $x = 1,\ 1.5,\ 2$. On the strip $1 \le x \le 3$, $y \in \mathbb{R}$, both $f$ and

are continuous and bounded, so $f$ is Lipschitz in $y$. By Picard’s theorem the IVP has a unique solution on $[1,3]$, and the standard convergence theory for one-step methods applies. Nothing is singular on the interval used.

Step 1 — The exact solution (needed for the comparison, and cheap)

The equation is linear of first order: $\dfrac{dy}{dx} - \dfrac{1}{x}y = 1$. The integrating factor is

so

The condition $y(1) = 1$ gives $1 = 0 + C$, i.e. $C = 1$. Hence

Check: $y' = 1 + \log x + 1 = 2 + \log x$, while $1 + y/x = 1 + (1+\log x) = 2 + \log x$. ✓

In particular

Note also $y'' = 1/x > 0$: the exact solution is convex on $[1,3]$. This is used in Step 4.

Step 2 — (i) Classical fourth-order Runge–Kutta, one step of size $h = 1$

The classical RK4 formulae are

With $x_0 = 1$, $y_0 = 1$, $h = 1$, so that $x_0 + \tfrac{h}{2} = 1.5$ and $x_0 + h = 2$:

$k_1$:

$k_2$: argument $\left(1.5,\ 1 + \tfrac{k_1}{2}\right) = (1.5,\ 2)$,

$k_3$: argument $\left(1.5,\ 1 + \tfrac{k_2}{2}\right) = \left(1.5,\ 1 + \tfrac{7}{6}\right) = \left(1.5,\ \tfrac{13}{6}\right)$,

$k_4$: argument $\left(2,\ 1 + k_3\right) = \left(2,\ \tfrac{31}{9}\right)$,

Combine. Over the common denominator $18$:

Therefore

Step 3 — (ii) Euler’s method

Euler: $y_{i+1} = y_i + h\,f(x_i, y_i)$, again with $h = 1$.

Step 1 ($x_0 = 1 \to x_1 = 2$):

Step 2 ($x_1 = 2 \to x_2 = 3$):

(Note the second Euler step is started from the computed value $y_1 = 3$, not from the exact $y(2)$ — this is why the error compounds.)

Step 4 — Comparison with the exact solution, and why RK4 wins

Relative errors at $x = 2$: Euler $11.41\%$, RK4 $0.20\%$ — a factor of about $58$ in accuracy for four function evaluations instead of one.

Why RK4 is so much closer. The local truncation errors are

so the global orders are $O(h)$ and $O(h^4)$. But note carefully: with $h = 1$ the powers of $h$ buy nothing — $h^2 = h^5 = 1$. The entire advantage here is structural, not asymptotic. Euler uses a single slope, taken at the left endpoint, and extrapolates it linearly across a whole unit interval; RK4 samples the slope field four times (at $x = 1,\ 1.5,\ 1.5,\ 2$) and weights the samples $\tfrac16(1,2,2,1)$ — Simpson’s weights — so that its increment reproduces the Taylor expansion of the exact solution through the $h^4$ term instead of only through $h^1$. It is the number of matched Taylor terms, not the smallness of $h$, that is doing the work at this step size.

Why both undershoot. Since $y'' = 1/x > 0$, the solution is convex, so the Euler local error

is positive at every step: each Euler step lands below the curve. Moreover the error is propagated forward with the factor $1 + h\,\partial f/\partial y = 1 + 1/x_i > 0$, so successive deficits reinforce rather than cancel. Euler must therefore undershoot at $x=2$ and undershoot by more at $x=3$ — which is exactly what the table shows ($3 < 3.386$, $5.5 < 6.296$). This is a prediction, not a post-hoc observation.

Remark 1 (the Euler iteration in closed form). With $h=1$ and $x_n = n+1$, Euler reads $y_{n+1} = y_n\dfrac{n+2}{n+1} + 1$. Putting $y_n = (n+1)z_n$ gives $z_{n+1} = z_n + \dfrac{1}{n+2}$ with $z_0 = 1$, so

Compare the exact $y(x_n) = (n+1)\bigl(1 + \log(n+1)\bigr)$: Euler has simply replaced $\log(n+1) = \int_1^{n+1}\frac{dt}{t}$ by its right-endpoint rectangle sum $\sum_{j=2}^{n+1}\frac1j$ on unit subintervals. Because $1/t$ is decreasing, that sum underestimates the integral — an independent proof of the undershoot. The exact global error is

$\gamma$ being Euler’s constant: the error grows linearly and without bound, confirming that $h=1$ is far too coarse for Euler. (Check: $n=1$ gives $2[1-(1.5-\log 2)] = 0.3862944$ ✓, $n=2$ gives $3[1-(\tfrac{11}{6}-\log 3)] = 0.7958369$ ✓.)

Remark 2 (RK4 continued to $x = 3$, for completeness — not asked). A second RK4 step from $\left(2,\tfrac{365}{108}\right)$ gives

an error of $0.0108369$ against $6.2958369$ — still $73$ times better than Euler at the same point. Under exam conditions do not compute this: part (i) asks for $y(2)$ only.

$\qquad\blacksquare$

Answer