Ordinary differential equations

Ordinary Differential Equations

Often in Physics, Chemistry, Biology, Geometry, etc there arise equations that relate a function with its derivative, or successive derivatives.

Definition - Ordinary differential equation. An ordinary differential equation (O.D.E.) is a equation that relates an independent variable $x$ , a function $y (x)$ that depends on $x$ , and the successive derivatives of $y$ , $y^{'}, y^{″}, \dots, y^{(n)}$ ; it can be written as

$F (x, y, y^{'}, y^{″}, \dots, y^{(n)}) = 0.$

The order of a differential equation is the greatest order of the derivatives in the equation.

Example. The equation $y^{‴} + s e n (x) y^{'} = 2 x$ is a differential equation of order 3.

Deducing a differential equation

To deduce a differential equation that explains a natural phenomenon is essential to understand what a derivative is and how to interpret it.

Example. Newton’s law of cooling states

“The rate of change of the temperature of a body in a surrounding medium is proportional to the difference between the temperature of the body $T$ and the temperature of the medium $T_{a}$ .”

The rate of change of the temperature is the derivative of temperature with respect to time $d T / d t$ . Thus, Newton’s law of cooling can be explained by the differential equation

$\frac{d T}{d t} = k (T - T_{a}),$

where $k$ is a proportionality constant.

Solution of an ordinary differential equation

Definition - Solution of an ordinary differential equation. Given an ordinary differential equation $F (x, y, y^{'}, y^{″}, \dots, y^{(n}) = 0$ , the function $y = f (x)$ is a solution of the ordinary differential equation if it satisfies the equation, that is, if

$F (x, f (x), f^{'} (x), f^{″} (x), \dots, f^{(n} (x)) = 0.$

The graph of a solution of the ordinary differential equation is known as integral curve.

Solving an ordinary differential equations consists on finding all its solutions in a given domain. For that integral calculus is required.

The same manner than the indefinite integral is a family of antiderivatives, that differ in a constant term, after integrating an ordinary differential equation we get a family of solutions that differ in a constant. We can get particular solutions giving values to this constant.

General solution of an ordinary differential equation

Definition - General solution of an ordinary differential equation. Given an ordinary differential equation $F (x, y, y^{'}, y^{″}, \dots, y^{(n}) = 0$ of order $n$ , the general solution of the differential equation is a family of functions

$y = f (x, C_{1}, \dots, C_{n}),$

depending on $n$ constants, such that for any value of $C_{1}, \dots, C_{n}$ we get a solution of the differential equation.

For every value of the constant we get particular solution of the differential equation. Thus, when a differential equation can be solved, it has infinite solutions.

Geometrically, the general solution of a differential equation corresponds to a family of integral curves of the differential equation.

Often, it is common to impose conditions to the solutions of a differential equation to reduce the number of solutions. In many cases, these conditions allow to determine the values of the constants in the general solution to get a particular solution.

First order differential equations

In this chapter we are going to study first order differential equations

$F (x, y, y^{'}) = 0.$

The general solution of a first order differential equation is

$y = f (x, C),$

so to get a particular solution from the general one, it is enough to set the value of the constant $C$ , and for that we only need to impose one initial condition.

Definition - Initial value problem. The group consisting of a first order differential equation and an initial condition is known as initial value problem:

${\begin{cases} F (x, y, y^{'}) = 0, & First order differential equation; \\ y (x_{0}) = y_{0}, & Initial condition. \end{cases}$

Solving an initial value problem consists in finding a solution of the first order differential equation that satisfies the initial condition.

Example. Recall the first order differential equation of the Newton’s law of cooling, $\frac{d T}{d t} = k (T - T_{a}),$ where $T$ is the temperature of the body and $T_{a}$ is the temperature of the surrounding medium.

It is easy to check that the general solution of this equation is

$T (t) = C e^{k t} + T_{a} .$

If we impose the initial condition that the temperature of the body at the initial instant is $5$ ºC, that is, $T (0) = 5$ , we have

$T (0) = C e^{k \cdot 0} + T_{a} = C + T_{a} = 5,$

from where we get $C = 5 - T_{a}$ , and this give us the particular solution

$T (t) = (5 - T_{a}) e^{k t} + T_{a} .$

Integral curve of an initial value problem

Example. If we assume in the previous example that the temperature of the surrounding medium is $T_{a} = 0$ ºC and the cooling constant of the body is $k = 1$ , the general solution of the differential equation is $T (t) = C e^{t},$ that is a family of integral curves. Among all of them, only the one that passes through the point $(0, 5)$ corresponds to the particular solution of the previous initial value problem.

Existence and uniqueness of solutions

Theorem - Existence and uniqueness of solutions of a first order ODE. Given an initial value problem

${\begin{cases} y^{'} = F (x, y); \\ y (x_{0}) = y_{0}; \end{cases}$

if $F (x, y (x))$ is a function continuous on an open interval around the point $(x_{0}, y_{0})$ , then a solution of the initial value problem exists. If, in addition, $\frac{\partial F}{\partial y}$ is continuous in an open interval around $(x_{0}, y_{0})$ , the solution is unique.

Although this theorem guarantees the existence and uniqueness of a solution of a first order differential equation, it does not provide a method to compute it. In fact, there is not a general method to solve first order differential equations, but we will see how to solve some types:

Separable differential equations
Homogeneous differential equations
Linear differential equations

Separable differential equations

Definition - Separable differential equation. A separable differential equation is a first order differential equation that can be written as

$y^{'} g (y) = f (x),$

or what is the same,

$g (y) d y = f (x) d x,$

so the different variables are on different sides of the equality (the variables are separated).

The general solution for a separable differential equation comes after integrating both sides of the equation

$\int g (y) d y = \int f (x) d x + C .$

Example. The differential equation of the Newton’s law of cooling

$\frac{d T}{d t} = k (T - T_{a}),$

is a separable differential equation since it can be written as

$\frac{1}{T - T_{a}} d T = k d t .$

Integrating both sides of the equation we have

$\int \frac{1}{T - T_{a}} d T = \int k d t \Leftrightarrow \log (T - T_{a}) = k t + C,$

and solving for $T$ we get the general solution of the equation

$T (t) = e^{k t + C} + T_{a} = e^{C} e^{k t} + T_{a} = C e^{k t} + T_{a},$

rewriting $C = e^{C}$ as an arbitrary constant.

Homogeneous differential equations

Definition - Homogeneous function. A function $f (x, y)$ is homogeneous of degree $n$ , if it satisfies

$f (k x, k y) = k^{n} f (x, y),$

for any value $k \in R$ .

In particular, a homogeneous function of degree $0$ always satisfies

$f (k x, k y) = f (x, y) .$

Setting $k = 1 / x$ we have

$f (x, y) = f (\frac{1}{x} x, \frac{1}{x} y) = f (1, \frac{y}{x}) = g (\frac{y}{x}) .$

This way, a homogeneous function of degree $0$ always can be written as a function of $u = y / x$ :

$f (x, y) = g (\frac{y}{x}) = g (u) .$

Definition - Homogeneous differential equation. A homogeneous differential equation is a first order differential equation that can be written as

$y^{'} = f (x, y),$

where $f (x, y)$ is a homogeneous function of degree $0$ .

We can solve a homogeneous differential equation by making the substitution

$u = \frac{y}{x} \Leftrightarrow y = u x,$

so the equation becomes

$u^{'} x + u = f (u),$

that is a separable differential equation.

Once solved the separable differential equation, the substitution must be undone.

Example. Let us consider the following differential equation $4 x - 3 y + y^{'} (2 y - 3 x) = 0.$

Rewriting the equation in this way

$y^{'} = \frac{3 y - 4 x}{2 y - 3 x}$

we can easily check that it is a homogeneous differential equation.

To solve this equation we have to do the substitution $y = u x$ , and we get

$u^{'} x + u = \frac{3 u x - 4 x}{2 u x - 3 x} = \frac{3 u - 4}{2 u - 3}$

that is a separable differential equation.

Separating the variables we have

$u^{'} x = \frac{3 u - 4}{2 u - 3} - u = \frac{- 2 u^{2} + 6 u - 4}{2 u - 3} \Leftrightarrow \frac{2 u - 3}{- 2 u^{2} + 6 u - 4} d u = \frac{1}{x} d x .$

Now, integrating both sides of the equation we have

$\begin{matrix} \int \frac{2 u - 3}{- 2 u^{2} + 6 u - 4} d u = \int \frac{1}{x} d x \Leftrightarrow - \frac{1}{2} \log | u^{2} - 3 u + 2 | = \log | x | + C \Leftrightarrow \\ \Leftrightarrow \log | u^{2} - 3 u + 2 | = - 2 \log | x | - 2 C, \end{matrix}$

then, applying the exponential function to both sides and simplifying we get the general solution

$u^{2} - 3 u + 2 = e^{- 2 \log | x | - 2 C} = \frac{e^{- 2 C}}{e^{\log | x |^{2}}} = \frac{C}{x^{2}},$

rewriting the constant $K = e^{- 2 C}$ .

Finally, undoing the initial substitution $u = y / x$ , we arrive at the general solution of the homogeneous differential equation

${(\frac{y}{x})}^{2} - 3 \frac{y}{x} + 2 = \frac{K}{x^{2}} \Leftrightarrow y^{2} - 3 x y + 2 x^{2} = K .$

Linear differential equations

Definition - Linear differential equation A linear differential equation is a first order differential equation that can be written as

$y^{'} + g (x) y = h (x) .$

To solve a linear differential equation we try to write the left side of the equation as the derivative of a product. For that we multiply both sides by the function $f (x)$ , such that

$f^{'} (x) = g (x) f (x) .$

Thus, we get

$\begin{matrix} y^{'} f (x) + g (x) f (x) y = h (x) f (x) \\ ⇕ \\ y^{'} f (x) + f^{'} (x) y = h (x) f (x) \\ ⇕ \\ \frac{d}{d x} (y f (x)) = h (x) f (x) \end{matrix}$

Integrating both sides of the previous equation we get the solution

$y f (x) = \int h (x) f (x) d x + C .$

On the other hand, the unique function that satisfies $f^{'} (x) = g (x) f (x)$ is

$f (x) = e^{\int g (x) d x},$

so, substituting this function in the previous solution we arrive at the solution of the linear differential equation

$y e^{\int g (x) d x} = \int h (x) e^{\int g (x) d x} d x + C,$

or what is the same

Solution of a linear differential equation.

$y = e^{- \int g (x) d x} (\int h (x) e^{\int g (x) d x} d x + C) .$

Example. If in the differential equation of the Newton’s law of cooling the temperature of the surrounding medium is a function of time $T_{a} (t)$ , then the differential equation

$\frac{d T}{d t} = k (T - T_{a} (t)),$

is a linear differential equation since it can be written as

$T^{'} - k T = - k T_{a} (t),$

where the independent term is $- k T_{a} (t)$ and the coefficient of $T$ is $- k$ .

Substituting in the formula of the general solution of a linear differential equation we have

$y = e^{- \int - k d t} (\int - k T_{a} (t) e^{\int - k d t} d t + C) = e^{k t} (- \int k T_{a} (t) e^{- k t} d t + C) .$

In the particular case that $T_{a} (t) = t$ , and the proportionality constant $k = 1$ , the general solution of the linear differential equation is

$y = e^{t} (- \int t e^{- k t} d t + C) = e^{t} (e^{- t} (t + 1) + C) = C e^{t} + t + 1.$

If, in addition, we know that the temperature of the body at time $t = 0$ is $5$ ºC, that is, we have the initial condition $T (0) = 5$ , then we can compute the value of the constant $C$ ,

$y (0) = C e^{0} + 0 + 1 = 5 \Leftrightarrow C + 1 = 5 \Leftrightarrow C = 4,$ and we get the particular solution

$y (t) = 4 e^{t} + t + 1.$

Ordinary Differential Equation

Last updated on Sep 25, 2020