5.15 Slope fieldsIB Maths: Applications and Interpretation HL: Revision notes

A slope field (direction field) for a first-order differential equation $\frac{dy}{dx}=f(x,y)$ is a grid of short line segments. At each point $(x,y)$ the segment has the gradient $f(x,y)$. Example: for $\frac{dy}{dx}=x-y$, the segment at $(3,1)$ has gradient $3-1=2$ and the segment at $(0,1)$ has gradient $-1$. A slope field lets you see the shape of solutions without solving the differential equation.

To use a slope field, substitute the coordinates of a point into $\frac{dy}{dx}$ to find the gradient there, then draw (or describe) the segment with that gradient: positive means rising, negative means falling, $0$ means horizontal. To draw a slope field, work out the gradient at a grid of points, for example integer values of $x$ and $y$, and draw a short segment through each point. Example: for $\frac{dy}{dx}=\frac{x}{y}$ the gradient at $(-3,6)$ is $-\frac12$.

An isocline is a curve on which all segments have the same gradient. Set $f(x,y)=c$ for a constant $c$. Horizontal segments occur where $f(x,y)=0$. For $\frac{dy}{dx}=x-y$, that is the line $y=x$. For $\frac{dy}{dx}=x+y$, the segments on $y=2-x$ all have gradient $2$, and those on $y=-x-1$ all have gradient $-1$. If $f(x,y)$ is undefined, for example $y=0$ in $\frac{dy}{dx}=\frac{x}{y}$, the segment is vertical or missing there.

A solution curve follows the segments: it is tangent to the segment at every point it passes through. The general solution is the whole family of such curves. An initial condition, such as passing through $(0,1)$, picks out one curve (the particular solution). Its tangent at the starting point has gradient $f(0,1)$. Solution curves do not cross, because the slope field gives only one gradient at each point. If a line is itself a solution, for example $y=-x-1$ in $\frac{dy}{dx}=x+y$, other curves stay on one side of it.

For a model such as $\frac{dP}{dt}=0.2P(5-P)$, horizontal segments at $P=0$ and $P=5$ show equilibrium values where the population is constant. Between $P=0$ and $P=5$ the gradient is positive, so the population increases. Above $P=5$ the gradient is negative, so it decreases. Solution curves therefore move towards $P=5$. At $P=1$, $\frac{dP}{dt}=0.8$ hundred fish per year; at $P=6$, $\frac{dP}{dt}=-1.2$. State what happens in the long term, with units and context.

When the equation is separable, you can check a slope field against the exact solution. For $\frac{dy}{dx}=\frac{x}{y}$, separation gives $y^2=x^2+C$. With $(1,2)$: $4=1+C$, so $C=3$ and $y=\sqrt{x^2+3}$. To show that a function is a solution, differentiate it and substitute it into the equation, as with $y=e^x-x-1$ for $\frac{dy}{dx}=x+y$. Use a GDC to evaluate a solution at a value, for example $y(2)=e^2-3=4.39$.