AHL 5.16—Eulers method for 1st order DEs

Here $f(x,y)=2x-y$, $x_0=0$, $y_0=3$, $h=0.1$.

$f(0,3)=2(0)-3=-3$.

$y_1=y_0+h\cdot f(x_0,y_0)=3+0.1(-3)=3-0.3=2.7$.

So $y(0.1)\approx 2.7$.

$3.3$ results from adding instead of subtracting the correction, i.e. a sign error on the slope.

$3.0$ ignores the correction term entirely, leaving $y$ unchanged.

$2.9$ would come from using $h=0.0333$ incorrectly or a slope of $-1$, not matching $f(0,3)=-3$.

Only $2.7$ follows correctly from $y_0+hf(x_0,y_0)$.

The $x$-update rule is $x_{n+1}=x_n+h$.

Here $x_0=1$ and $h=0.5$.

$x_1=1+0.5=1.5$.

The $x$-value after one step is $1.5$.

$1.0$ is simply the original $x_0$, showing no step was taken.

$2.0$ would result from doubling $x_0$ incorrectly rather than adding $h$.

$0.5$ is just the step size itself, mistaken for the new $x$-value.

Only $x_1=x_0+h=1.5$ is correct.