Ordinary Differential Equation (ODE):

Definition

An ordinary differential equation (ODE) involves an unknown function $y(x)$ of a single variable $x$ and its derivatives. It can be represented as:

$$F(x, y, y', y'', \ldots, y^{(n)}) = 0$$

where $y' = \frac{dy}{dx}$ represents the first derivative of $y$ with respect to $x$, and $y'', y''', \ldots, y^{(n)}$ denote the second, third, and $n$th derivatives of $y$, respectively.

Solving an ODE involves finding the functional form of $y(x)$ that satisfies the given differential equation, subject to certain initial or boundary conditions. Analytical techniques, such as separation of variables, variation of parameters, and integrating factors, are employed to solve ODEs and obtain explicit solutions.

Order and degree

To determine the degree of the ODE, we consider the highest power of the highest derivative in the equation. For example, if the highest derivative is \(y''\), and it appears with a power of 2 or higher, the degree is 2. However, if the highest derivative appears with a power of 1, the degree is 1.

The order of the ODE corresponds to the highest derivative present in the equation. For example, if the equation contains only the first derivative \(y'\), the order is 1. If the equation contains the second derivative \(y''\) but not higher derivatives, the order is 2.

Solutions

solving a second order ODE of degree 1

A second-order ordinary differential equation (ODE) of degree 1 can be solved using various methods depending on the specific form of the equation. One common approach is to convert the second-order ODE into a system of first-order ODEs, also known as a first-order ODE system, and then solve it using techniques such as the initial value problem or boundary value problem methods.

Let's consider a second-order ODE of degree 1 in the form:

$$y''(x) + p(x)y'(x) + q(x)y(x) = r(x)$$

To solve this equation, we can introduce new variables to convert it into a system of first-order ODEs. Let's define \(v(x) = y'(x)\), which gives us the following:

$$y'(x) = v(x) $$

$$y''(x) = v'(x)$$

Substituting these expressions into the original equation, we obtain:

$$v'(x) + p(x)v(x) + q(x)y(x) = r(x)$$

Now, we have a system of two first-order ODEs:

$$y'(x) = v(x) $$

$$v'(x) = -p(x)v(x) - q(x)y(x) + r(x)$$

We can solve this system of first-order ODEs using numerical methods or analytical techniques, such as the method of integrating factors, separation of variables, or variation of parameters.

The specific method of solving the first-order ODE system depends on the coefficients \(p(x)\), \(q(x)\), and the function \(r(x)\) in the original second-order ODE. By solving the first-order ODE system, we can obtain the solutions \(y(x)\) and \(v(x)\), which correspond to the original second-order ODE.

It's important to note that the general solution of a second-order ODE contains two arbitrary constants, typically determined by the initial conditions or boundary conditions associated with the problem.