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