Expert Answer to "Solve differential equation cos2⁡y4x+2dy=(cos⁡y+sin⁡y)2x2+x+3dx"

Name: Solve differential equation cos2⁡y4x+2dy=(cos⁡y+sin⁡y)2x2+x+3dx
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve differential equation cos2⁡y4x+2dy=(cos⁡y+sin⁡y)2x2+x+3dx

College
Calculus and Analysis
Differential Equations
First order differential equations

Chaya Galloway

Answered question

2021-03-07

Solve differential equation $\frac{\cos^{2} y}{4 x + 2} d y = \frac{(\cos y + \sin y)^{2}}{\sqrt{x^{2} + x + 3}} d x$

Answer & Explanation

mhalmantus

Skilled2021-03-08Added 105 answers

$\frac{\cos^{2} y}{4 x + 2} d y = \frac{\cos^{2} y + \sin^{2} y + 2 \sin y \cos y}{\sqrt{x^{2} + x + 3}} d x$
$\frac{\cos^{2} y}{4 x + 2} d y = \frac{1 + \sin 2 y}{\sqrt{x^{2} + x + 3}} d x$
$\frac{\cos 2 y}{1 + \sin 2 y} d y = \frac{4 x + 2}{\sqrt{x^{2} + x + 3}} d x$
Integrating on both sides
$\int \frac{\cos 2 y}{1 + \sin 2 y} d y = \int \frac{4 x + 2}{\sqrt{x 2 + x + 3}} d x$
$\int \frac{4 x + 2}{\sqrt{x 2 + x + 3}} d x$
$L e t x^{2} + x + 3 = u^{2}$
$2 x + 1 d x = 2 u d u$
$4 x + 2 d x = 4 u d u$
$\int \frac{4 x + 2}{\sqrt{x 2 + x + 3}} d x = \frac{4 u d u}{u} d u$
$= 4 d u = 4 u + c$
$= 4 \sqrt{x^{2} + x + 3} + c$
$\int \frac{\cos 2 y}{1 + \sin 2 y} y$
$L e t t = 1 + \sin 2 y$
Differentiating
$d t = 2 \cos 2 y$
$\int \frac{\cos 2 y}{1 + \sin 2 y} d y = \int \frac{\frac{1}{2} d t}{t}$
$= \frac{1}{2} \int \frac{1}{t} d t$ $= \frac{1}{2} \ln t + K$
$\frac{1}{2} \ln (1 + \sin 2 y) + k$
$\int \frac{\cos 2 y}{1 + \sin 2 y} d y = \int \frac{4 x + 2}{\sqrt{x 2 + x + 3}} d x$
$4 \sqrt{x 2 + x + 3} = \frac{1}{2} \ln (1 + \sin 2 y) + c$

Jeffrey Jordon

Expert2021-12-25Added 2605 answers

Answer is given below (on video)

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