AHL 5.15—Slope fields

The slope at any point is given by $f(x,y)=x+y$.

Substitute $x=2$, $y=-3$ into $f(x,y)=x+y$.

$f(2,-3)=2+(-3)=-1$.

The slope of the segment at $(2,-3)$ is $-1$.

$1$ would result from computing $x-y=2-(-3)=5$ incorrectly or a sign slip; it does not match $x+y$.

$5$ comes from computing $x-y=2-(-3)=5$, mistakenly subtracting instead of adding.

$-5$ comes from computing $-(x-y)$, a double sign error combining subtraction and negation.

Only direct substitution into $x+y$ gives $-1$, confirming option A.

Set $y(3-y)=0$, giving $y=0$ and $y=3$.

For $0<y<3$, e.g. $y=1$: $f(1)=1(2)=2>0$, so curves rise toward $y=3$.

For $y<0$, e.g. $y=-1$: $f(-1)=-1(4)=-4<0$, so curves fall away from $y=0$.

For $y>3$, e.g. $y=4$: $f(4)=4(-1)=-4<0$, so curves fall back toward $y=3$.

Curves move away from $y=0$ (unstable) and toward $y=3$ from both sides (stable).

This reverses the actual behaviour; it wrongly assumes flow moves toward $y=0$ and away from $y=3$.

This assumes both equilibria are attracting, but $y=0$ has curves diverging away from it below and above.

This assumes both equilibria repel, ignoring that curves in $0<y<3$ and $y>3$ both approach $y=3$.

Only option A correctly matches the sign analysis: $y=0$ unstable, $y=3$ stable.