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 $(2)$

$\blacksquare$

Sources

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