Direction fields and solution curves

What you will be able to do

• find the slope that an equation gives at a point, and draw a short segment;
• read a direction field and find where the slope is zero or equal;
• take a short step along the field and compare it with the exact solution.

A map of slopes

Suppose $\dfrac{dy}{dx} = x - y$. At the point $(2, 1)$ the slope is $2 - 1 = 1$. At $(1, 3)$ the slope is $1 - 3 = -2$. At every point of the plane the equation names a slope. We do not need a solution to know it.

A direction field (also called a slope field) is a picture of these slopes. At each point of a grid we draw a short segment with the slope the equation gives. A solution curve is a curve that touches the segments as it goes, like a boat carried by a current. The curve through a point is the solution that starts there.

Reading the field

• Where the segments are flat, the slope is zero. For $\dfrac{dy}{dx} = x - y$ this happens along the line $y = x$.
• Below that line $x - y$ is positive, so the segments rise. Above it they fall.
• A solution curve cannot cross another one. At a crossing point the two curves would give two different slopes, but the equation gives only one.
• All the curves approach the line $y = x - 1$ as $x$ grows. You can see this in the picture, and you can check that $y = x - 1$ solves the equation: its slope is $1$, and $x - y = x - (x - 1) = 1$.

Isoclines

An isocline is a curve along which the slope has the same value. For $\dfrac{dy}{dx} = x - y$ the isocline for slope $m$ is the line $x - y = m$, that is, $y = x - m$. On it every segment is parallel. If you draw a few isoclines, you can sketch the whole field. Slide $m$ in the simulation and watch the dashed line move.

An exact solution to compare

The equation $\dfrac{dy}{dx} = x - y$ is linear, and you will solve it in the next course. Its solutions are $y = x - 1 + Ke^{-x}$. The curve through $(0, 1)$ has $1 = -1 + K$, so $K = 2$, and the curve is $y = x - 1 + 2e^{-x}$. Differentiate to check: $y' = 1 - 2e^{-x}$, and $x - y = 1 - 2e^{-x}$. They agree.

Walking along the field

Start at $(0, 1)$. The slope there is $0 - 1 = -1$. Move $0.5$ to the right along that slope. The height changes by $-1 \times 0.5$, so the new point is $(0.5, 0.5)$. Repeat from there. This is the idea of Euler's method: new height equals old height plus slope times step.

A short step stays close to the true curve. A long step drifts away. The course on modelling uses the method on real problems.

Practise with new numbers

Try it in Python

The program walks along the field of $\dfrac{dy}{dx} = x - y$ from $(0, 1)$ in four steps of $0.5$. Next to each stop it prints the exact value from $y = x - 1 + 2e^{-x}$.

Common mistakes

• Using the slope as the height. The equation gives the slope. It does not give $y$.
• Drawing a field with crossing solutions. Two solution curves never cross, because one point has one slope.
• A step that is too long. A long step along a bending curve leaves the curve.
• Mixing up the axes. At the point $(x, y)$ the slope $x - y$ uses the first number as $x$.

Remember these

• The equation $\dfrac{dy}{dx} = f(x, y)$ gives a slope at every point. A direction field draws them.
• A solution curve follows the segments. Curves do not cross.
• An isocline is a curve of equal slope.
• A step along the field is $y_{\text{new}} = y_{\text{old}} + h \times f(x_{\text{old}}, y_{\text{old}})$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. For $\dfrac{dy}{dx} = x^2 - y$, find the slope at $(2, 1)$.

2. For $\dfrac{dy}{dx} = x - y$, what is the slope along the line $y = x - 2$?

3. For $\dfrac{dy}{dx} = y(2 - y)$, which horizontal lines are solutions?

4. For $\dfrac{dy}{dx} = x + y$ with the curve through $(0, 1)$, take one step of $h = 0.1$.

5. Why can two solution curves of $\dfrac{dy}{dx} = f(x, y)$ not cross?

References

• OpenStax, Calculus Volume 2, Chapter 4, section 4.2 Direction Fields and Numerical Methods.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the introduction and the section on solutions of a differential equation.