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

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## 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 $(2)$ |

$\blacksquare$

## Sources

- 1963: Morris Tenenbaum and Harry Pollard:
*Ordinary Differential Equations*... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: Example $3.52$