Solution to Differential Equation/Examples
Examples of Solutions to Differential Equations
Arbitrary Order $1$ ODE: $1$
Consider the real function defined as:
- $y = \ln x + C$
defined on the domain $x \in \R_{>0}$.
Then $\map f x$ is a solution to the first order ODE:
- $(1): y' = \dfrac 1 x$
defined on the domain $x \in \R_{>0}$.
Arbitrary Order $1$ ODE: $2$
Consider the real function defined as:
- $y = \tan x - x$
defined on the domain $S := \set {x \in \R: x \ne \dfrac {\paren {2 n + 1} \pi} 2, n \in \Z}$.
Then $\map f x$ is a solution to the first order ODE:
- $(1): y' = \paren {x + y}^2$
when $x$ is restricted to $S$.
Arbitrary Order $2$ ODE
Consider the real function defined as:
- $y = \map f x = \ln x + x$
defined on the domain $x \in \R_{>0}$.
Then $\map f x$ is a solution to the second order ODE:
- $(1): \quad x^2 y'' + 2 x y' + y = \ln x + 3 x + 1$
defined on the domain $x \in \R_{>0}$.
Arbitrary Order $2$, Degree $3$ ODE
Consider the equation:
- $(1): \quad y = x^2$
where $x \in \R$.
Then $(1)$ is a solution to the second order ODE:
- $(2): \quad \paren {y''}^3 + \paren {y'}^2 - y - 3 x^2 - 8 = 0$
defined on the domain $x \in \R$.
Equation which is Not a Solution
Consider the equation:
- $(1): \quad y = \sqrt {-\paren {1 + x^2} }$
where $x \in \R$.
Consider the first order ODE:
- $(2): \quad x + y y' = 0$
Then despite the fact that the formal substition for $y$ and $y'$ from $(1)$ into $(2)$ yields an identity, $(1)$ is not a solution to $(2)$.
Absolute Value Function
Consider the real function defined as:
- $\map f x = \size x$
where $\size x$ is the absolute value function.
Then $\map f x$ cannot be the solution to a differential equation.
However, by suitably restricting $\map f x$ to a domain which does not include $x = 0$, there may well exist differential equations for which the resulting real function is a solution.