Answered: "y′−sin⁡2x+ytan⁡x=0"

Jerold

Answered question

2021-08-31

$y^{'} - \sin 2 x + y \tan x = 0$

Answer & Explanation

lamusesamuset

Skilled2021-09-01Added 93 answers

Solve the linear equation $\frac{d y (x)}{d x} - \sin (2 x) + \tan (x) y (x) = 0$ :
Add $\sin (2 x)$ to both sides:
$\frac{d y (x)}{d x} + \tan (x) y (x) = \sin (2 x)$
Let $μ (x) = e^{\int \tan (x) d x} = \sec (x)$ .
Multiply both sides by $μ (x)$ :
$\sec (x) \frac{d y (x)}{d x} + (\sec (x) \tan (x)) y (x) = \sec (x) \sin (2 x)$
INTERMEDIATE STEPS:
Take the integral:
$\int \tan (x) d x$
Rewrite $\tan (x) a s \frac{\sin (x)}{\cos (x)}$ :
$\frac{\sin (x)}{\cos (x)}$
For the integrand $\frac{\sin (x)}{\cos (x)}$ , substitute $u = \cos (x) and d u = - \sin (x) d x$ :
$= \int - \frac{1}{u} d u$
Factor out constants:
$= \int - \frac{1}{u} d u$
The integral of $\frac{1}{u}$ is $\log (u)$ :
$= - \log (u) + constant$
Substitute back for $u = \cos (x)$ :
Answer:
$= - \log (\cos (x)) + constant$
Substitute $\tan (x) \sec (x) = \frac{d \sec (x)}{d x}$ :
$\sec (x) \frac{d y (x)}{d x} + \frac{d \sec (x)}{d x} y (x) = \sec (x) \sin (2 x)$
INTERMEDIATE STEPS:
Possible derivation:
$\frac{d}{d x} (\sec (x))$
Rewrite the expression:

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