Worked examples — Runge-Kutta 4th order (RK4) — derivation

This deep-dive drills the parent RK4 derivation into practice. We solve one step of $y^{\prime }=f(x,y)$ using the four slopes $k_1,k_2,k_3,k_4$ and the update $y_{n+1}=y_n+\frac{1}{6}(k_1+2k_2+2k_3+k_4)$. Nothing new is assumed — every symbol below was built in the parent note. Here we make sure no case surprises you.

Before any numbers, let's recall the machine we are feeding, so a reader arriving cold can follow.

Why re-list it? Because 90% of RK4 mistakes are plugging the wrong $x$ or $y$ into a stage. Keep this box open while you read.

Because our promise is visual-first, every worked example below comes with its own step-diagram: a picture of the slope samples on the actual curve, so the arithmetic never floats free of geometry. Look at the figure first, then read the numbers.

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The scenario matrix

Every RK4 problem you will ever see is one (or a mix) of the cells below. The examples that follow are labelled with the cell they hit, so together they cover the whole grid.

Cell	What makes it different	Example
A. Pure-$x$ integrand $f=f(x)$	RK4 collapses into Simpson's rule	Ex 1
B. Linear mixed $f=x+y$	Both variables move; smooth growth	Ex 2
C. Negative / decaying slope $f=-y$	Signs flip; $y$ shrinks, must not overshoot below	Ex 3
D. Nonlinear in $y$ $f=y^2$ etc.	Stages differ a lot; sensitivity	Ex 4
E1. Zero slope everywhere $f\equiv 0$	Nothing moves; sanity floor	Ex 5a
E2. Zero only at start $k_1=0$, rest $\ne 0$	Motion still happens; don't stop early	Ex 5b
F. Multi-step march (two steps of $h$)	Errors chain; $y_2$ from $y_1$	Ex 6
G. Large step / limiting $h$ big	Where accuracy strains; Taylor view	Ex 7
H. Word problem (real world) cooling / motion	Translate a story into $f$, keep units	Ex 8
I. Exam twist $f$ depends on $x$ and $y$ nonlinearly, awkward numbers	Careful arithmetic, no calculator luxuries	Ex 9

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Cell A — pure-$x$ integrand (RK4 = Simpson)

Forecast: The exact answer is ${\int }_0^1{x}^{2}dx=\frac{1}{3}$. RK4 should nail it exactly — guess why before reading on. (Hint: Simpson is exact on cubics.)

Figure 1 (Cell A): the parabola $y=\frac{x^3}{3}$ with the three sample $x$-values (left, midpoint, right) marked; because $f$ ignores $y$, the two midpoint samples coincide.

1. $k_1=hf(0)=1\cdot 0^2=0$. Why this step? Left-edge slope. Since $f$ ignores $y$, we just plug $x=0$.
2. $k_2=1\cdot f(0.5)=0.25$. Why? Midpoint $x=0+\frac{1}{2}$. The $y$-shift $y_n+\frac{k_1}{2}$ is irrelevant because $f$ has no $y$.
3. $k_3=1\cdot f(0.5)=0.25$. Why? Same midpoint, same value — with no $y$-dependence, $k_2=k_3$ always for pure-$x$ problems.
4. $k_4=1\cdot f(1)=1$. Why? Right edge $x=1$.
5. $y_1=0+\frac{1}{6}(0+2(0.25)+2(0.25)+1)$. The bracket is $0+0.5+0.5+1=2$, so $y_1=\frac{2}{6}=\frac{1}{3}.$

Verify: Exact ${\int }_0^1{x}^{2}dx=\frac{1}{3}$. RK4 error $=0$. This is Cell A's whole point: when $f=f(x)$, RK4 is Simpson's rule $\frac{h}{6}(f_0+4f_{mid}+f_1)$ and is exact on polynomials up to degree 3. See Simpson's Rule.

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Cell B — linear mixed slope

Forecast: Exact solution is $y=2e^x-x-1$, so $y(0.2)=2e^{0.2}-1.2\approx 1.2428$. Predict RK4 matches to ~5 decimals.

Figure 2 (Cell B): the true curve $y=2e^x-x-1$ from $x=0$ to $0.2$, with the four slope samples $k_1..k_4$ as short line segments at their sample points — see how each successive segment tilts a little steeper.

1. $k_1=0.2f(0,1)=0.2(0+1)=0.2$. Why? Left slope.
2. $k_2=0.2f(0.1,1+0.1)=0.2(0.1+1.1)=0.24$. Why? Midpoint $x=0.1$, $y$ shifted by $\frac{k_1}{2}=0.1$.
3. $k_3=0.2f(0.1,1+0.12)=0.2(0.1+1.12)=0.244$. Why? Refined midpoint, $y$ shifted by $\frac{k_2}{2}=0.12$.
4. $k_4=0.2f(0.2,1+0.244)=0.2(0.2+1.244)=0.2888$. Why? Right edge, $y$ shifted by full $k_3$.
5. $y_1=1+\frac{1}{6}(0.2+2(0.24)+2(0.244)+0.2888)=1+\frac{1.4568}{6}=\mathrm{1.2428.}$

Verify: $2e^{0.2}-1.2=1.242806\dots$; our $1.24280$ matches. Error $\sim 10^{-6}$ — textbook 4th-order behaviour.

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Cell C — negative / decaying slope

Forecast: Exact is $y=e^{-x}$, so $y(0.5)=e^{-0.5}\approx 0.6065$. The naive worry: could RK4 overshoot below zero? Guess: no, it stays positive.

Figure 3 (Cell C): the decaying curve $y=e^{-x}$; every slope sample points downward (negative), but the arrows get shorter as $y$ drops — the picture of a self-limiting decay that never dives below zero.

1. $k_1=0.5\cdot (-1)=-0.5$. Why? Slope $-y=-1$ at start; the negative sign says "go down".
2. $k_2=0.5\cdot -\left(1+\frac{-0.5}{2}\right)=0.5\cdot -(1-0.25)=0.5(-0.75)=-0.375$. Why? Midpoint $y$ is lower than start, so the downward slope is gentler.
3. $k_3=0.5\cdot -(1+\frac{-0.375}{2})=0.5\cdot -(1-0.1875)=0.5(-0.8125)=-0.40625$. Why? Refined midpoint.
4. $k_4=0.5\cdot -(1+(-0.40625))=0.5\cdot -(0.59375)=-0.296875$. Why? Right edge $y$ has dropped further.
5. $y_1=1+\frac{1}{6}(-0.5+2(-0.375)+2(-0.40625)+(-0.296875))=1+\frac{-2.359375}{6}=\mathrm{0.606771.}$

Verify: $e^{-0.5}=0.606531$; error $\approx 2\times 10^{-4}$ (larger step, still tiny). Crucially $y_1>0$ — RK4's averaged slope tracks the decaying curve rather than firing straight down like Euler (which would give $1-0.5=0.5$, way low). Compare with Euler's Method.

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Cell D — nonlinear in $y$

Forecast: Exact is $y=\frac{1}{1-x}$ (it blows up at $x=1$!). So $y(0.1)=\frac{1}{0.9}=1.11111$. Predict RK4 close to that.

Figure 4 (Cell D): the curve $y=\frac{1}{1-x}$ steepening toward its blow-up at $x=1$; the four slope arrows visibly lengthen from $k_1$ to $k_4$ because squaring amplifies each $y$-shift.

1. $k_1=0.1\cdot 1^2=0.1$. Why? Slope $y^2=1$.
2. $k_2=0.1\cdot (1+0.05{)}^2=0.1(1.1025)=0.11025$. Why? Midpoint $y=1.05$; squaring amplifies the shift.
3. $k_3=0.1\cdot (1+0.055125{)}^2=0.1(1.055125{)}^2=0.1(1.113289)=0.111329$. Why? Refined midpoint $y=1.0551$.
4. $k_4=0.1\cdot (1+0.111329{)}^2=0.1(1.111329{)}^2=0.1(1.235052)=0.123505$. Why? Right edge $y=1.1113$.
5. $y_1=1+\frac{1}{6}(0.1+2(0.11025)+2(0.111329)+0.123505)=1+\frac{0.666663}{6}=\mathrm{1.111111.}$

Verify: $\frac{1}{0.9}=1.111111$; error $\sim 10^{-7}$. Even near a singularity (far away here), RK4 shines because it senses the steepening slope through the midpoint refinements.

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Cell E — zero / degenerate inputs (two flavours)

Forecast: With zero slope, $y$ cannot change. Answer should be exactly $7$.

1. $k_1=2\cdot 0=0$. Why? Slope is $0$ everywhere.
2. $k_2=2\cdot 0=0$. Why? Still zero even after any $y$-shift, since $f\equiv 0$.
3. $k_3=0$, $k_4=0$. Why? Same reasoning — nothing to sample.
4. $y_1=7+\frac{1}{6}(0+0+0+0)=7.$

Verify: Exact $y\equiv 7$. Error $=0$. Sanity floor: if the true solution is constant, RK4 leaves it constant.

Forecast: Exact is $y=5+\frac{x^2}{2}$, so $y(1)=5.5$. The trap: seeing $k_1=0$ and wrongly concluding "$y$ doesn't change". It does — later stages sample nonzero slopes.

Figure 5 (Cell E2): the parabola $y=5+\frac{x^2}{2}$; the left-edge arrow $k_1$ is perfectly flat ($=0$), yet the midpoint and right-edge arrows tilt up — the curve climbs from $5$ to $5.5$ despite the flat start.

1. $k_1=1\cdot f(0)=1\cdot 0=0$. Why? Slope $=x=0$ at the left edge — genuinely flat here.
2. $k_2=1\cdot f(0.5)=1\cdot 0.5=0.5$. Why? Midpoint $x=0.5$; the $y$-shift is irrelevant since $f$ has no $y$, but $x$ has moved so the slope is now $0.5$.
3. $k_3=1\cdot f(0.5)=0.5$. Why? Same midpoint $x$; equals $k_2$ (pure-$x$ integrand).
4. $k_4=1\cdot f(1)=1$. Why? Right edge $x=1$, slope $=1$.
5. $y_1=5+\frac{1}{6}(0+2(0.5)+2(0.5)+1)=5+\frac{3}{6}=\mathrm{5.5.}$

Verify: Exact $5+\frac{1}{2}=5.5$. Error $=0$ (pure-$x$, so Simpson-exact). Lesson: never stop marching just because $k_1=0$ — always compute all four stages.

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Cell F — multi-step march

Forecast: Same ODE as Ex 2 but reached in two small steps. Two steps of $h=0.1$ should beat one step of $h=0.2$ (Ex 2 result $1.242806$). Predict $y_2\approx 1.24281$ and slightly more accurate.

Figure 6 (Cell F): two consecutive RK4 steps stitched along $y=2e^x-x-1$; the first step lands at $(0.1,1.11034)$, and the second starts from that point to reach $(0.2,1.24281)$ — the picture of errors chaining step to step.

Step 1 (from the parent note's Worked Example 1): $y_1=1.11034$ at $x=0.1$. Why reuse? Identical arithmetic to the parent; we start the second step from here.

Step 2 (from $x=0.1$, $y=1.11034$, $h=0.1$):

1. $k_1=0.1f(0.1,1.11034)=0.1(0.1+1.11034)=0.121034$. Why? New left edge.
2. $k_2=0.1f(0.15,1.11034+0.060517)=0.1(0.15+1.170857)=0.132086$. Why? Midpoint $x=0.15$.
3. $k_3=0.1f(0.15,1.11034+0.066043)=0.1(0.15+1.176383)=0.132638$. Why? Refined midpoint.
4. $k_4=0.1f(0.2,1.11034+0.132638)=0.1(0.2+1.242978)=0.144298$. Why? Right edge $x=0.2$.
5. $y_2=1.11034+\frac{1}{6}(0.121034+2(0.132086)+2(0.132638)+0.144298)=1.11034+0.132463=\mathrm{1.242806.}$

Verify: Exact $2e^{0.2}-1.2=1.242806$. Two small steps land essentially bang-on. Lesson: halving $h$ shrinks global error by $\approx 2^4=16\times$ — the "4th order" promise (see Local vs Global Truncation Error).

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Cell G — large step / limiting behaviour

Forecast: Exact is $e=2.71828$. Euler would give $2$. RK4 with a huge step — guess it still gets within ~1%.

1. $k_1=1\cdot 1=1$. Why? Slope $=y=1$.
2. $k_2=1\cdot (1+0.5)=1.5$. Why? Midpoint $y=1.5$.
3. $k_3=1\cdot (1+0.75)=1.75$. Why? Refined midpoint.
4. $k_4=1\cdot (1+1.75)=2.75$. Why? Right edge $y=2.75$.
5. $y_1=1+\frac{1}{6}(1+3+3.5+2.75)=1+\frac{10.25}{6}=\mathrm{2.708333.}$

Verify: $e=2.718282$; error $\approx 0.0099$. And the beautiful identity: $y_1=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}=2.708\overline{3},$ the first five terms of $e^x=\sum \frac{x^n}{n!}$ at $x=1$. RK4 reproduces the Taylor series through the $h^4$ term exactly — this is why it is "4th order" (compare Taylor Series Methods). The missing piece $\frac{1}{120}+\cdots \approx 0.0099$ is precisely the $O(h^5)$ local error, visible because $h=1$ is large.

Figure 7 (Cell G): the true $e^x$ (dashed navy) against Euler's single left-slope line (orange, landing at $2$) and RK4's four sampled slopes (magenta dots) that bend the path up to $2.708$ — a labelled snapshot of why sampling many slopes wins.

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Cell H — real-world word problem

Forecast: Coffee cools toward $20^{\circ }$. Exact: $T=20+70e^{-0.1t}$, so $T(2)=20+70e^{-0.2}=20+57.31=77.31^{\circ }\text{C}$. Predict RK4 close to $77.3$.

Figure 8 (Cell H): the cooling curve $T=20+70e^{-0.1t}$ dropping from $90^{\circ }$ toward the dashed room-temperature line at $20^{\circ }$; the four slope samples show cooling slowing as $T$ approaches the room — the curve never crosses below $20^{\circ }$.

Here $f(t,T)=-0.1(T-20)$, units ${}^{\circ }\text{C}/\text{min}$; each $k_i$ has units ${}^{\circ }\text{C}$ (slope $\times$ minutes).

1. $k_1=2\cdot (-0.1)(90-20)=2(-7)=-14$. Why? Left slope: hot coffee cools fast.
2. $k_2=2(-0.1)((90-7)-20)=2(-0.1)(63)=-12.6$. Why? Midpoint $T=90+\frac{k_1}{2}=83$.
3. $k_3=2(-0.1)((90-6.3)-20)=2(-0.1)(63.7)=-12.74$. Why? Refined midpoint $T=83.7$.
4. $k_4=2(-0.1)((90-12.74)-20)=2(-0.1)(57.26)=-11.452$. Why? Right edge $T=77.26$.
5. $T_1=90+\frac{1}{6}(-14+2(-12.6)+2(-12.74)+(-11.452))=90-12.6887=\mathrm{77.3113.}$

Verify: Exact $20+70e^{-0.2}=77.3128^{\circ }\text{C}$; error $\approx 0.0015^{\circ }$. Units check: every $k_i$ is in ${}^{\circ }\text{C}$, the update adds ${}^{\circ }\text{C}$ to ${}^{\circ }\text{C}$ ✓. Physically sound: temperature dropped but stayed well above room's $20^{\circ }$.

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Cell I — exam-style twist

Forecast: Slope is positive (both $x,\sqrt{y}>0$), so $y$ grows. Rough guess: $y^{\prime }\approx 1\cdot 2=2$, so over $0.2$ minutes $y$ rises ~$0.4$-ish beyond simple estimate — expect $y\approx 4.4$.

Figure 9 (Cell I): the exact solution $y={\left(\frac{x^2}{4}+\frac{7}{4}\right)}^2$ climbing from $(1,4)$ to $(1.2,4.4527)$; the four slope arrows (all upward) grow because both $x$ and $\sqrt{y}$ increase across the step.

1. $k_1=0.2\cdot 1\cdot \sqrt{4}=0.2(1)(2)=0.4$. Why? Left edge $x=1$, $y=4$.
2. $k_2=0.2\cdot 1.1\cdot \sqrt{4+0.2}=0.2(1.1)\sqrt{4.2}=0.22(2.04939)=0.450866$. Why? Midpoint $x=1.1$, $y=4.2$.
3. $k_3=0.2\cdot 1.1\cdot \sqrt{4+0.225433}=0.22\sqrt{4.225433}=0.22(2.055586)=0.452229$. Why? Refined midpoint $y=4.22543$.
4. $k_4=0.2\cdot 1.2\cdot \sqrt{4+0.452229}=0.24\sqrt{4.452229}=0.24(2.110030)=0.506407$. Why? Right edge $x=1.2$, $y=4.45223$.
5. $y_1=4+\frac{1}{6}(0.4+2(0.450866)+2(0.452229)+0.506407)=4+\frac{2.712597}{6}=\mathrm{4.452100.}$

Verify: Separable exact solution: $\int y^{-1/2}dy=\int xdx\Rightarrow 2\sqrt{y}=\frac{x^2}{2}+C$. At $x=1,y=4$: $2(2)=\frac{1}{2}+C\Rightarrow C=\frac{7}{2}$. At $x=1.2$: $2\sqrt{y}=\frac{1.44}{2}+3.5=4.22\Rightarrow \sqrt{y}=2.11\Rightarrow y=4.4521$. Our RK4 gives $4.45210$ — matches to 4 decimals. Forecast of $\approx 4.4$ was in the right ballpark. ✓

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The matrix, all green

Every cell A–I now has a fully worked, verified example (with E split into E1 and E2). Sign flips (C), zeros both total and partial (E1, E2), nonlinearity (D, I), chaining (F), stress steps (G), and a units-bearing story (H) are all covered — you should never meet an RK4 scenario this page did not rehearse.

Recall Self-test before you close

Which cell tests that RK4 = Simpson? ::: Cell A (pure-$x$ integrand, Ex 1). Why must you keep computing when $k_1=0$? ::: Later stages ($k_2,k_3,k_4$) may sample nonzero slopes — see Ex 5b where $y$ climbs $5\to 5.5$ despite $k_1=0$. Why did the coffee stay above $20^{\circ }$? ::: The slope $-0.1(T-20)\to 0$ as $T\to 20$; cooling slows near room temperature. In Ex 7, what do $1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}$ represent? ::: The first five Taylor terms of $e^1$ — proof RK4 matches Taylor through $h^4$. Halving $h$ improves global error by what factor? ::: About $2^4=16\times$ (4th-order convergence).

Connections

• Euler's Method — the undershooting single-slope baseline (Ex 3, Ex 7).
• Modified Euler / Heun's Method — 2nd-order sibling for comparison.
• Simpson's Rule — exactly what Cell A / Ex 1 reduces to.
• Taylor Series Methods — the target series Ex 7 reproduces.
• Local vs Global Truncation Error — the $16\times$ rule in Ex 6.
• Adaptive Step Size (RKF45) — how to pick $h$ automatically for cells like G.