Calculus

1. Euler's method

Euler's method is a numerical method used to approximate the solution of a first-order ordinary differential equation (ODE) of the form \( \frac{dy}{dx} = f(x,y)\), given an initial condition \(y(x_0) = y_0\). It involves approximating the solution at a set of discrete points and using these approximations to approximate the solution over the entire range of \(x\).

The formula for Euler's method is:

$$y_{i+1} = y_i + hf(x_i,y_i)$$

where \(h\) is the step size and is given by \(h = \frac{b - a}{n}\), where \(a\) and \(b\) are the limits of the interval over which the solution is desired, \(n\) is the number of subintervals, and \(x_i\) and \(y_i\) represent the values of \(x\) and \(y\) at the \(i^{th}\) subinterval.

For example, let us use Euler's method to approximate the solution of the differential equation \( \frac{dy}{dx} = x + y\) with initial condition \(y(0) = 1\) over the interval \([0, 1]\) with a step size of \(h = 0.1\):

First, we compute the values of x and y at each subinterval using the formula:

$$y_{i+1} = y_i + hf(x_i,y_i)$$

Starting with \(x_0 = 0\) and \(y_0 = 1\), we have:

$$y_1 = y_0 + hf(x_0,y_0) = 1 + 0.1(0 + 1) = 1.1$$ $$y_2 = y_1 + hf(x_1,y_1) = 1.1 + 0.1(0.1 + 1.1) = 1.22$$ $$y_3 = y_2 + hf(x_2,y_2) = 1.22 + 0.1(0.2 + 1.22) = 1.353$$

Continuing in this way, we can compute the values of y at each subinterval. The approximation of the solution of the differential equation is the set of ordered pairs \( {(x_0, y_0), (x_1, y_1), (x_2, y_2), ..., (x_n, y_n)} \).

If the question had been "Approximate the value of \(y\) when \(x=0.3\)" the answer would be \(y_3\) which is \(1.353\).

Euler's method can be used to approximate the solution of a wide range of first-order ODEs, but its accuracy depends on the step size h. Smaller step sizes generally lead to more accurate approximations, but at the cost of increased computation time.

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You can use the TInSpire to calculate using Euler's Method with these Instructions:

Using the calculator's spreadsheet :: video

Using the calculator's function :: video (Not available in Press to Test)

Using the calculator's graph plotter :: video (Not available in Press to Test)

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2. Directly integrable

Consider very simple differential equations where \(\frac{dy}{dx}\) (the rate of change of \(y\) with respect to \(x\)) is given as an easily integrable function of \(x\).

For instance, consider the differential equation \(\frac{dy}{dx} = 5x\). Here, the rate of change of \(y\) is directly proportional to \(x\). To solve this, we integrate both sides with respect to \(x\).

$$ \int \frac{dy}{dx} \, dx = \int 5x \, dx \\ \implies y = \frac{5}{2}x^2 + c $$

Where \(c\) is a constant of integration. This equation represents the family of functions that satisfy the given differential equation.

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3. Variables Separable

A differential equation of the variables separable type is a type of differential equation that can be rewritten by algebraic operations into a form that separates the variables on two sides of the equation, i.e., \(g(y) dy = h(x) dx\). The solution involves integrating both sides of the equation.

The general steps to solve a separable differential equation are:

$$ \begin{align*} 1. & \quad \text{Rewrite the equation in the form } g(y) dy = h(x) dx \\\ 2. & \quad \text{Integrate both sides to obtain } \int g(y) dy = \int h(x) dx \\\ 3. & \quad \text{Solve for } y \text{ to find the general solution} \end{align*} $$

Now, let's see an example:

Solve the following separable differential equation: \(\frac{dy}{dx} = x e^y\)

The solution proceeds as follows:

$$ \begin{align*} 1. & \quad \text{Rewrite the equation in the form } e^{-y} dy = x dx \\\ 2. & \quad \text{Integrate both sides: } \int e^{-y} dy = \int x dx \\\ 3. & \quad \text{This yields } -e^{-y} = \frac{1}{2}x^2 + c \\\ 4. & \quad \text{Finally, solving for } y, \text{ we get: } y = - \ln\left( -\frac{x^2}{2} - c \right) \end{align*} $$

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4. Homogeneous

A homogeneous differential equation is a special type of differential equation in which each term is a homogeneous function of the same degree. A first-order homogeneous differential equation can be written in the form \( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \). To solve this equation, we typically use the substitution \( y = vx \), which simplifies the differential equation into a separable form.

The substitution leads to two relationships: \( y = vx \) and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substituting these into the original equation results in a new equation in terms of \( v \) and \( x \) that can be solved using separation of variables.

The key formulae are:

$$ \begin{align*} 1. & \quad y = vx \\\ 2. & \quad \frac{dy}{dx} = v + x\frac{dv}{dx} \\ \end{align*} $$

Now, let's see an example:

Solve the following homogeneous differential equation: \( \frac{dy}{dx} = \frac{y^2 - x^2}{xy} \)

The solution proceeds as follows:

$$ \begin{align*} 1. & \quad \text{Set } y = vx, \text{ hence } \frac{dy}{dx} = v + x\frac{dv}{dx} \\\ 2. & \quad \text{Substitute } y = vx \text{ and } \frac{dy}{dx} = v + x\frac{dv}{dx} \text{ into the original equation:} \\\ & \quad v + x\frac{dv}{dx} = \frac{(vx)^2 - x^2}{x(vx)} = \frac{v^2x^2 - x^2}{vx^2} = v - \frac{1}{v} \\\ 3. & \quad \text{This simplifies to a separable equation: } x\frac{dv}{dx} = - \frac{1}{v} \\\ 4. & \quad \text{Separating variables and integrating, we have: } \\\ & \quad \int v dv = \int - \frac{1}{x} dx \\\ 5. & \quad \text{Solving these integrals, we find: } \\\ & \quad \frac{1}{2} v^2 = - \ln |x| + C \\\ 6. & \quad \text{Finally, substituting } v = \frac{y}{x} \text{ back in, we get: } \\\ & \quad \frac{y^2}{2x^2} = - \ln |x| + C \\\ & \quad y = \sqrt{2x^2(C - \ln |x|) }\\ \end{align*} $$

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5. Integrating factor

Consider a first-order linear differential equation of the form \(y' + P(x)y = Q(x)\). To solve this type of equation, we often use a technique called the integrating factor method. The key to this method is to multiply the equation by an "integrating factor", which is the exponential of the integral of the function \(P(x)\), i.e., \(e^{\int P(x) dx}\). This transformation converts the left-hand side of the equation into the derivative of a product, making it possible to solve the equation.

The key formula for the integrating factor method is as follows:

$$ \begin{align*} \quad \text{The integrating factor is: } e^{\int P(x) dx} \\\ \end{align*} $$

Now, let's see an example:

Solve the following differential equation: \(y' + 2xy = x\)

The solution proceeds as follows:

$$ \begin{align*} 1. & \quad \text{Identify } P(x) = 2x \text{ and } Q(x) = x \\\ 2. & \quad \text{The integrating factor is } e^{\int 2x dx} = e^{x^2} \\\ 3. & \quad \text{Multiply the entire equation by the integrating factor: } e^{x^2}y' + 2x e^{x^2}y = xe^{x^2} \\\ 4. & \quad \text{The left-hand side is now the derivative of } (y e^{x^2}), \text{ so: } (y e^{x^2})' = xe^{x^2} \\\ 5. & \quad \text{Integrate both sides to get: } y e^{x^2} = \int xe^{x^2} dx + C \\\ 6. & \quad \text{This integral is a standard result: } \frac{1}{2}e^{x^2} + C \\\ 7. & \quad \text{Finally, solving for } y(x), \text{ we get: } y(x) = \frac{\frac{1}{2}e^{x^2} + C}{e^{x^2}} = \frac{1}{2} + Ce^{-x^2} \end{align*} $$

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6. Maclaurin series

In some cases the differential equation \( \frac{dy}{dx} = f(x, y) \) with initial condition \( (0,y_0) \) can be solved by expressing the differential as an infinite polynomial. Provided the necessary derivatives exist, we can determine the nth degree Maclaurin polynomial approximation by finding \(y', y'', y''' \) and then creating the series as described in Syllabus Content 5.19

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