# 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.