Expert Answer to "Linear equations of first order.Solve the initial-value problem..."

High School
Algebra
Algebra I
Forms of linear equations

Rui Baldwin

Answered question

2021-01-31

Linear equations of first order.
Solve the initial-value problem on the specified interval $y^{'} + y \tan x = \sin 2 x o n (- \frac{1}{2} π, \frac{1}{2} π) with y = 2 when x = 0$ .

Answer & Explanation

bahaistag

Skilled2021-02-01Added 100 answers

From the given linear differential equation, $P (x) = \tan x$ and $Q (x) = \sin 2 x .$
Therefore, the integrating factor is given by
$(x) = \int_{0}^{x} \tan t d t = {[- \ln \cos t]}_{0}^{x} = - \ln \cos x$ .
So that, the solution is given by
$y = 2 \cos x + \cos x \int_{1}^{x} {(\cos t)}^{- 1} \sin 2 t h t$
$= 2 \cos x + \cos x \int_{1}^{x} {(\cos t)}^{- 1} (2 \sin t \cos t) d t$
$= 2 \cos x + \cos x \int_{1}^{x} \sin t d t$
$= 2 \cos x + \cos x {[- \cos t]}_{0}^{x}$
$= 2 \cos x + \cos x (1 - \cos x)$
$= 4 \cos x - 2 \cos^{2} x$ .
We can verify the function is the solution of the initial-value periblem since it satisfies the differential equation and the initial condition.

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