First Order Differential Equations

The Euler Method

The Euler method is a numerical technique for approximating solutions to first-order differential equations. It's like taking baby steps along a curve, using the derivative to guide each step.

Consider a point $(x_n, y_n)$, where the derivative of this point is $$y_n'(x_n, y_n)$$. The tangent line at this point is given by

$$y - y_n= y_n'(x_n, y_n)(x-x_n)$$

$$y = y_n + y_n'(x_n, y_n)(x-x_n)$$

Now let us consider a new point $x_{n+1}, y_{n+1}$, where $x_{n+1}$ is defined by $x_{n+1}= x_{n}+h$ for some small $h$.

Here comes the approximation step. We can imagine that, for some small $h$, the rate of change at the points $(x_n, y_n)$ and $(x_{n+1}, y_{n+1})$ are incredibly similar to the point where they are effectively the same.

If the rate of change at these points are similar, the point $(x_{n+1}, y_{n+1})$ should lie on the tangent line of $(x_n, y_n)$

Therefore, by substituting the point $(x_{n+1}, y_{n+1})$ into the tangent line:

$$y_{n+1} = y_n + y_n'(x_n,y_n)(x_{n+1}-x_n)$$

$$y_{n+1} = y_n + y_n'(x_n)(h)$$

$$y_{n+1} = y_n + hy_n'(x_n)$$

Or alternatively, by letting $f(x_n, y_n) = y_n'(x_n,y_n)$

$$y_{n+1} = y_n + h\cdot f(x_n, y_n)$$

Where:

$y_{n+1}$ is the next y-value
$y_n$ is the current y-value
$h$ is the step size
$f(x_n, y_n)$ is the derivative at the current point

Thus obtaining a sequence of values $(x_n, y_n)$ that satisfy a differential equation.

Tip

Choose a smaller step size ($h$) for better accuracy, but be aware this increases computational work. For smaller $h$, the approximation of rate of changes being similar between $x_n$ and $x_{n+1}$ becomes more accurate.

Example

Solve $\frac{dy}{dx} = x + y$ with initial condition $y(0) = 1$ using h = 0.2 for the first two steps.

Step 1: At x = 0, y = 1 $y_1 = 1 + 0.2(0 + 1) = 1.2$

Step 2: At x = 0.2, y = 1.2 $y_2 = 1.2 + 0.2(0.2 + 1.2) = 1.48$

Tip

Remember that the Euler's method is not foolproof. There are cases when it doesn't work, for example.

When the rate of change increases rapidly such that the derivative approximation is invalid.
Asymptotes—certain differential equations don't have solutions at certain points, but Euler's method neglects this and spits out a number anyway.

Tip

For using this with a GDC, go to "Recursion" and define the equations:

$$x_{n+1}=x_n + h$$

$$y_{n+1}=y_n + hf(x_n,y_n)$$

Along with the initial condition (i.e. the values of $x_0$ and $y_0$ .

The calculator will do the rest.

Variables Separable Method

This method works when we can separate variables to different sides of the equation.

Steps:

Rearrange to get all $y$ terms on one side and $x$ terms on the other
Integrate both sides

Example

Solve $\frac{dy}{dx} = \frac{x}{y}$

Separate variables: $y,dy = x,dx$
Integrate: $\frac{y^2}{2} = \frac{x^2}{2} + C$
Solution: $y = \pm\sqrt{x^2 + 2C}$

Integrating Factor Method

Used when the DE is in the form: $$\frac{dy}{dx} + P(x)y = Q(x)$$

The integrating factor is: $$ \mu(x) = e^{\int P(x)dx}$$

The reason why this works is because if we multiply both sides of the DE by the integrating factor:

$$ e^{\int P(x) dx}\frac{dy}{dx}+yP(x)e^{\int P(x) dx}=e^{\int P(x) dx}Q(x) $$

We can notice that this can now be written in the form:

$$\frac{d}{dx}(ye^{\int P(x)dx} ) = e^{\int P(x) dx}Q(x)$$

Therefore: $$ye^{\int P(x)dx}=\int e^{\int P(x) dx}Q(x) dx$$

Which hopefully, is a solve-able integral that you can do.

Tip

Look for terms with y and dy/dx that can be combined into a product rule form.

Example

Solve $\frac{dy}{dx} + 2xy = x$

Identify $P(x) = 2x$ and $Q(x) = x$
Find $\mu(x) = e^{\int 2x,dx} = e^{x^2}$
Multiply both sides: $e^{x^2}\frac{dy}{dx} + 2xe^{x^2}y = xe^{x^2}$
Recognize left side as derivative: $\frac{d}{dx}(e^{x^2}y) = xe^{x^2}$
Integrate: $e^{x^2}y = \frac{1}{2}e^{x^2} + C$
Solve for y: $y = \frac{1}{2} + Ce^{-x^2}$

Homogeneous Differential Equations

A DE is homogeneous if it can be written as: $$\frac{dy}{dx} = f(\frac{y}{x})$$

Solve using substitution $y = vx$, where $v$ is a function of $x$:

Let $y = vx$
Use substitution $\frac{dy}{dx} = v + x\frac{dv}{dx}$
Substitute and solve for v

Common Mistake

Don't forget to substitute back y = vx after solving for v!

Example

Solve $\frac{dy}{dx} = \frac{y}{x} + \frac{y^2}{x^2}$

Let $y = vx$
$v + x\frac{dv}{dx} = v + v^2$
$x\frac{dv}{dx} = v^2$
Separate: $\frac{dv}{v^2} = \frac{dx}{x}$
Integrate: $-\frac{1}{v} = \ln|x| + C$
Solve for v and substitute back: $y = \frac{-x}{\ln|x| + C}$

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Differential Equation

A differential equation is an equation that involves an unknown function and its derivatives. It describes how a quantity changes with respect to another.

The order of a differential equation is determined by the highest derivative present.
The solution to a differential equation is a function, not a number.

ExampleFor example, $\frac{dy}{dx} = 2x$ is a first-order differential equation because it involves the first derivative of $y$ with respect to $x$ .

AnalogyThink of a differential equation as a recipe that tells you how to cook a dish (the function) by specifying the rate at which you should add ingredients (the derivatives).

DefinitionDifferential EquationAn equation involving an unknown function and its derivatives.

Number and Algebra15 subtopics

Functions10 subtopics

Geometry and Trigonometry16 subtopics

Statistics and Probability19 subtopics

Calculus18 subtopics

AHL 5.14—Setting up a DE, solve by separating variables Notes

Number and Algebra15 subtopics

Functions10 subtopics

Geometry and Trigonometry16 subtopics

Statistics and Probability19 subtopics

Calculus18 subtopics

First Order Differential Equations

The Euler Method

Variables Separable Method

Integrating Factor Method

Homogeneous Differential Equations

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Differential Equation

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