# Solution to Differential Equation/Examples/Arbitrary Order 1 ODE: 2

## Examples of Solutions to Differential Equations

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 $S$.

## Proof

It is noted that $\map f x$ is not defined in $\R$ when $x = \dfrac {\paren {2 n + 1} \pi} 2$ because for those values of $x$ the tangent is not defined.

Hence the restriction of $x$ to $S$

It is also noted that $(1)$ is indeed defined for all $x \in S$.

Having established that, we continue:

 $\ds y$ $=$ $\ds \tan x - x$ $\text {(2)}: \quad$ $\ds \leadsto \ \$ $\ds y'$ $=$ $\ds \sec^2 x - 1$ Derivative of Tangent Function, Derivative of Identity Function

For ease of manipulation we rewrite $(1)$ as:

$(2): y' - \paren {x + y}^2 = 0$

Then:

 $\ds$  $\ds \paren {\sec^2 x - 1} - \paren {x + \tan x - x}^2$ substituting for $y$ and $y'$ from above into the left hand side of $(2)$ $\ds$ $=$ $\ds \tan^2 x - \paren {\tan x}^2$ Difference of Squares of Secant and Tangent $\ds$ $=$ $\ds 0$ which equals the right hand side of $(2)$

$\blacksquare$