Lesson 1 - What Are Differential Equations? What Should I Know Before Beginning?

Understanding Notation

In Calculus, we learn about n $^{t h}$ order derivatives. An n $^{t h}$ is written as such:

\frac{d^{n} y}{d x^{n}}, y^{(n)} (x); n \geq 1

Here, $n \geq 1$ is an integer, $y$ is the dependent variable, and $x$ is the independent variable.

Consider $u (x, y)$ . Then $u_{x}$ or $\frac{\partial u}{\partial x}$ means partial derivative of $u$ with respect to $x$ .

A differential equation, or a DE, is an equation that contains any derivatives - either ordinary derivatives or partial derivatives.

Let's practice classifying derivatives.

a) $m \frac{d v}{d t} = F (t, v)$
b) $\frac{d y}{d x} + P (x) y = Q (x)$
c) $a y^{″} + b y^{'} + c y = g (t)$
d) $c o s (y) y^{″} (x) = (1 - y) y^{'} + y + 10$
e) $y^{(3)} (t) - y^{″} (t) - y = 10 x + e^{9 - y}$

See answers at the end of this section.

Differential equations can also be classified by order. The order of a DE is the largest derivative present in the differential equation. Try to classify the above equations by order.

Moreover, we can classify a differential equation by their derivatives as ordinary (ODE) or partial (PDE).

Finally, for the purposes of this course, we can classify a DE as linear or non-linear. A linear DE is a differential equation that can be written in the form:

a_{n} y^{(n)} (t) + a_{n - 1} y^{(n - 1)} (t) + . . . + a_{1} y^{'} (t) + a_{0} = g (t)

A DE that is not of this form is considered non-linear.

Solutions of Differential Equations

A solution to the DE can either be explicit or implicit.

An explicit solution is a solution that is written as the dependent variable in terms of the independent variable.

Example

a) Prove $y = s i n x + x^{2}$ is a explicit solution of $y^{″} + y = x^{2} + 2$ .

Solution

We begin by identifying the solution and the differential equation. We know that an equation is a differential equation because it contains derivatives. Thus, we can determine that $y^{″} + y = x^{2} + 2$ is the differential equation.

The goal is more algebraic than it seems. We simply are testing to ensure that the left side of the equation is equal to the right side of the equation. How can we do this?

Well we have two unknown variables in our solution equation: $y^{″}$ and $y$ . In this instance, $y$ is fortunately given to us.

y = s i n x + x^{2}

y^{″} + (s i n x + x^{2}) = x^{2} + 2

Now that we have plugged in our $y$ equation, we must find $y^{″}$ . To do so, we turn to our Calculus knowledge.

y = s i n (x) + x^{2}

\frac{d}{d x} (y) = \frac{d}{d x} (s i n (x) + x^{2})

\frac{d y}{d x} = c o s (x) + 2 x

We can say that $\frac{d y}{d x} = y^{'}$ . Now that we have found the first derivative, we must take the second derivative ( $y^{″}$ ).

\frac{d y}{d x} = c o s (x) + 2 x

\frac{d^{2} y}{d x^{2}} = - s i n (x) + 2

Now, we have all of our unknowns. Let's see if they match up.

y^{″} = - s i n (x) + 2

(- s i n (x) + 2) + (s i n x + x^{2}) = x^{2} + 2

Notice that the $- s i n (x)$ and $s i n (x)$ cancel out. We are left with

2 + x^{2} = x^{2} + 2 ✓

We can also have solutions to equations that are implicit. Those, we would have to solve for one variable in terms of the other in order to check for equivalency, if possible. Otherwise, we recall implicit differentiation from our Calculus toolbook.

b) Check if $x^{2} - s i n (x + y) = 1$ is a solution of $y^{'} = 2 x s e c (x + y) - 1$

We implement implicit differentiation.

\frac{d}{d x} (x^{2} - s i n (x + y)) = \frac{d}{d x} (1)

2 x - c o s (x + y) (1) + y^{'} = 0

Check to see if you can manipulate this equation to match the given equation by solving for $y^{'}$ .

- \cos (x + y) \times 1 + y^{'} = - 2 x

1 + y^{'} = - \frac{2 x}{\cos (x + y)}

y^{'} = - 2 x \times \frac{1}{\cos (x + y)} - 1

Recall inverse trig functions.

y^{'} = - 2 x \sec (x + y) - 1 ✓

Initial Value Problem (IVP)

Now, we sometimes we are given an initial condition in what is known as an Initial Value Problem.

Let's consider $\frac{d y}{d x} = y, y (1) = 2$ .

In order to solve the IVP, we need to find the General Solution to the equation.

\frac{d y}{d x} = y

\frac{1}{y} d y = d x

\int \frac{1}{y} = \int d x

\ln y = x + C

e^{\ln y} = e^{x + C}

y = C e^{x}

Now that we have a general solution, we can plug in the initial value given and solve for C.

y (1) = 2

2 = C e^{1}

C = \frac{2}{e}

Now that we have C, we have a particular solution to the IVP. The particular solution not only satisfied the original differential equation, but also satisfies the initial condition.

y = \frac{2}{e} e^{x}

Next Lesson: Lesson 2 - Existence and Uniqueness Theorem

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