Solve-dy-dx-ysec-x-tan-x-

Question Number 17735 by tawa tawa last updated on 09/Jul/17

Solve : \frac{dy}{dx} + ysec (x) = \tan (x)

Answered by alex041103 last updated on 10/Jul/17

So {let}^{'} s find function μ (x) wich satisfies the followinv

\frac{dy}{dx} μ + P (x) μ y = \frac{d}{dx} [y μ]

We know that

\frac{d}{dx} [y μ] = \frac{dy}{dx} μ + y \frac{d μ}{dx}

\Rightarrow P (x) μ = \frac{d μ}{dx}

Now solving the saparable differential

equation

P (x) dx = \frac{1}{μ} d μ

\Rightarrow \int P (x) dx = \int \frac{1}{μ} d μ

or \ln ∣ μ ∣= \int P (x) dx (we {won}^{'} t worry about the constant)

\Rightarrow μ (x) = e^{\int P (x) dx}

Now we apply the μ (x) to solve the differential

equation we started with .

P (x) = \sec x

\Rightarrow μ (x) = e^{\int \sec x dx}

We know the trivial result \int \sec x dx

= \ln ∣ \sec x + \tan x ∣ (+ C)

\Rightarrow μ (x) = \sec x + \tan x

\Rightarrow \tan x (\sec x + \tan x) = \frac{d}{dx} [y (\sec x + \tan x)]

\Rightarrow y (x) = \frac{\int \tan x \sec x dx + \int \tan^{2} x dx}{\sec x + \tan x}

We solve the integrals an we get

y (x) = 1 - \frac{x + C}{\sec x + \tan x}

Commented by alex041103 last updated on 10/Jul/17

\int \tan (x) \sec (x) dx =

= \int \frac{\sin (x)}{\cos^{2} (x)} dx

Let u = \cos (x) \Rightarrow du = - \sin (x)

\Rightarrow \int \tan (x) \sec (x) dx =

= - \int u^{- 2} du = \frac{1}{u} + C = \sec (x) + C

Commented by tawa tawa last updated on 10/Jul/17

Wow, God bless you sir .

Commented by alex041103 last updated on 10/Jul/17

\int \tan^{2} (x) dx =

= \int (\sec^{2} (x) - 1) dx

= \int \sec^{2} (x) dx - \int dx

= \tan (x) - x + C