Question

Q1.Find the solution of Differential equation.


y sin(2x)dx−(1+y²+cos²x)dy=0
​

Answer 1

Step-by-step explanation:

FORMULA TO BE IMPLEMENTED

1. If u & v are two differentiable function then

$d(uv) = v \: du + u \: dv</p><p></p><p></p><p>[tex]\displaystyle \int\limits_{}^{} x^{n} \, dx = \frac{ {x}^{n + 1} }{n + 1} + constant </p><p>$

EVALUATION

$y \sin 2x \: dx - ( 1 + {y}^{2} + { \cos}^{2}x )dy = 0$

Multiplying both sides by (-1) we get

$- y \sin 2x \: dx + ( 1 + {y}^{2} + { \cos}^{2}x )dy = 0$

Rearranging we get

$( 1 + {y}^{2} )dy + ( { \cos}^{2}x \: dy \: - y \sin 2x \: dx ) = 0$

$\implies \: ( 1 + {y}^{2} )dy + \bigg[ { \cos}^{2}x \: dy \: + y \times ( - 2\sin x \cos x\: )dx \bigg ] = 0$

On integration

$\displaystyle \int\limits_{}^{} ( 1 + {y}^{2} )dy +\displaystyle \int\limits_{}^{} d(y \: { \cos}^{2}x ) = 0 </p><p>$

$\implies \: \displaystyle \: y + \frac{ {y}^{3} }{3} + y \: { \cos}^{2} x \: = c$

Where c is a constant

RESULT

SO THE REQUIRED SOLUTION IS

$\boxed{ \: \displaystyle \: y + \frac{ {y}^{3} }{3} + y \: { \cos}^{2} x \: = c} </p><p>$

Answer 2

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