Recipe

Take one RK4 step

Use when: you're given an IVP $y'=f(t,y)$ , $y(t_0)=y_0$ , a chosen step size $h$ , and asked to advance the numerical solution by one step with fourth-order accuracy

Goal: the next point $(t_{n+1},y_{n+1})$ on the numerical solution

Taught in RK4

Quick recipe

Compute $k_1$ , the slope at the start
Compute $k_2$ , a first estimate at the midpoint
Compute $k_3$ , a second, better midpoint estimate
Compute $k_4$ , the slope at the endpoint
Combine all four with the $1{:}2{:}2{:}1$ weighting

1
$k_1=f(t_n,y_n)$ — evaluate the differential equation exactly where you already stand.

Example

For $y'=-2ty$ , $y_0=1$ at $t_0=0$ : $k_1=f(0,1)=-2(0)(1)=0$ .

Why: $k_1$ is the ordinary Euler slope — the differential equation evaluated at a point that's known exactly, with no estimation involved yet.
2
$k_2=f\!\left(t_n+\tfrac h2,\ y_n+\tfrac h2k_1\right)$ — walk half a step using $k_1$ , then ask $f$ for the slope there.

Example

$t_0+\tfrac h2=0.25$ , $y_0+\tfrac h2k_1=1+0.25(0)=1$ . $k_2=f(0.25,1)=-2(0.25)(1)=-0.5$ .
- $f$ does not depend on $t$ (an autonomous equation) — $t_n+\tfrac h2$ never enters the calculation at all; see the shortcut below.
- $f$ depends on both $t$ and $y$ — compute $t_n+\tfrac h2$ explicitly before evaluating $f$ there, as in the example.
Why: $k_2$ is a first, Euler-quality guess at the slope in the MIDDLE of the step — better-centered information than $k_1$ alone provides.
3
$k_3=f\!\left(t_n+\tfrac h2,\ y_n+\tfrac h2k_2\right)$ — the SAME midpoint, reached using $k_2$ instead of $k_1$ .

Example

$y_0+\tfrac h2k_2=1+0.25(-0.5)=0.875$ . $k_3=f(0.25,0.875)=-2(0.25)(0.875)=-0.4375$ .

⚡ Shortcut

For an AUTONOMOUS equation, $f$ ignores $t$ entirely, so this step (and Step 2) never need $t_n+\tfrac h2$ tracked at all — only the $y$ -argument changes between them. Section 6's own $y'=y$ example: $k_1=f(y_0)=1$ , $k_2=f(y_0+\tfrac h2k_1)=f(1.25)=1.25$ , $k_3=f(y_0+\tfrac h2k_2)=f(1.3125)=1.3125$ — no $t$ -bookkeeping anywhere, only $y$ -values.

Why: $k_2$ was only Euler-quality; refining the walk to the midpoint using $k_2$ instead of $k_1$ gives a second, more trustworthy midpoint slope before the step ever reaches the endpoint.
4
$k_4=f(t_n+h,\ y_n+hk_3)$ — walk the FULL step using $k_3$ 's refined midpoint slope, then ask $f$ for the slope there.

Example

$t_0+h=0.5$ , $y_0+hk_3=1+0.5(-0.4375)=0.78125$ . $k_4=f(0.5,0.78125)=-2(0.5)(0.78125)=-0.78125$ .

Why: $k_4$ estimates the slope at the FAR end of the step, using the best midpoint information ( $k_3$ ) available to get there — completing the set of four points the method samples before committing to a step.
5
$y_{n+1}=y_n+\dfrac h6(k_1+2k_2+2k_3+k_4)$ — NOT a plain average; the two midpoint estimates each count double.

Example

$y_1=1+\dfrac{0.5}{6}\big(0+2(-0.5)+2(-0.4375)+(-0.78125)\big)=1+0.08333(-2.65625)\approx0.77865$ . (True value $e^{-0.25}\approx0.77880$ — error $\approx0.00015$ .)

Why: this exact, non-uniform weighting is what makes the combined step match the true solution's Taylor expansion through the $h^4$ term — see rk4-formula's own proof for why $1{:}2{:}2{:}1$ specifically achieves that match.

Common mistakes

Using $t_n$ instead of $t_n+\tfrac h2$ when evaluating $k_2$ or $k_3$ — both midpoint stages need the SAME shifted $t$ -argument, not the step's own starting time.
Treating the weighting as a plain average — dividing by $4$ with equal weights (instead of dividing by $6$ with weights $1{:}2{:}2{:}1$ ) silently drops the method back toward lower accuracy.
Computing $k_4$ from $k_2$ instead of $k_3$ — $k_4$ needs the SECOND (refined) midpoint slope to step across the full interval, not the first.
Forgetting that $k_2$ , $k_3$ , and $k_4$ all depend on the PREVIOUS stage's own computed value — none of the four can be computed out of order.