ExerciseODEsintermediate

Exercise: Euler and RK4 by hand on a first-order ODE

Solve for y'=f(t,y) in an ODE given in implicit form and approximate y(3) with two Euler steps and one RK4 step, comparing both results.

Setup

Consider the ODE $y+(t^2y-t)\,y'=0$ with $y(1)=2$ , and approximate $y(3)$ . The equation is not in normal form, so first solve for $y'$ :

y'=f(t,y)=\frac{-y}{t^2y-t}

Two Euler steps

ExampleEuler with h=1

Apply two steps of explicit Euler with $h=1$ from $t_0=1$ , $y_0=2$ .

First step: at $(1,2)$ the denominator is $1^2\cdot 2-1=1$ , so $f(1,2)=-2/1=-2$ .
$y_1=y_0+h\,f(1,2)=2+1\cdot(-2)=0$
Second step: at $(2,0)$ the numerator is $-0=0$ , hence $f(2,0)=0$ and the solution does not change.
$y_2=y_1+h\,f(2,0)=0+0=0$

Euler with $h=1$ gives $y(3)\approx 0$ . The step is too large: the first jump takes the solution to $y=0$ , a point where $f$ vanishes and the method stays.

One Runge-Kutta step

ExampleRK4 with h=2

Apply one step of classical Runge-Kutta with $h=2$ from $t_0=1$ , $y_0=2$ .

Initial slope: $k_1=f(1,2)=-2$ .
Midpoint ( $t=2$ ) advancing with $k_1$ : $y=2+1\cdot(-2)=0$ , so $k_2=f(2,0)=0$ .
Midpoint again, now with $k_2$ : $y=2+1\cdot 0=2$ , so $k_3=f(2,2)=\frac{-2}{4\cdot 2-2}=-\frac{1}{3}$ .
Final endpoint ( $t=3$ ) with $k_3$ : $y=2+2\cdot(-\tfrac13)=\tfrac43$ , so $k_4=f\bigl(3,\tfrac43\bigr)=\frac{-4/3}{9\cdot\frac43-3}=-\frac{4}{27}$ .
Combination with weights $1,2,2,1$ :
$y_1=2+\frac{2}{6}\Bigl(-2+2\cdot 0+2\cdot\bigl(-\tfrac13\bigr)-\tfrac{4}{27}\Bigr)=2-\frac{76}{81}=\frac{86}{81}$

RK4 gives $y(3)\approx\frac{86}{81}=1.0617$ . With the same number of evaluations as four Euler steps, it avoids the collapse at $y=0$ and produces a reasonable approximation.