Math, asked by BrainlyPopularman, 7 months ago

Find the solution of Differential equation.

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

Answers

Answered by Anonymous
110

ANSWER :

\implies \sf \dfrac {-1}{2} \: y \: cos \: 2x \: - \: \dfrac {3y}{2} \: - \: \dfrac {y^{3}}{3} \: = \: 0

GIVEN :

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

SOLUTION :

\sf pdx \: + \: Qdx \: = \: 0

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

\sf \dfrac {dp}{dy} \: = \: sin \: 2x

\sf \dfrac {dQ}{dx} \: = \: - 2 \: cos \: x \times (-2 \: sin \: x) \: = \: 2 \: sin \: x \: cos \: x

\sf \int pdx \: = \: \dfrac {1}{2} y \: cos \: 2x \: + \:+ \: g(y)

\sf \int Qdy \: = \: - y \: - \: \dfrac {y^{33}}{3} \: - \: y \: cos^{2}x \: + \: h(x)

\sf cos^{2}x \: = \: \dfrac {1}{2} \: (1 \: + \: cos \: 2x)

\sf g(y) \: - \: y \: - \: \dfrac {y^{3}}{3} \: - \: \dfrac {y}{2} \: and \: h(x) \: = \: 0

\sf \dfrac {-1}{2} \: y \: cos \: 2x \: - \: \dfrac {3y}{2} \: - \: \dfrac {y^{3}}{3}

\implies \sf \dfrac {-1}{2} \: y \: cos \: 2x \: - \: \dfrac {3y}{2} \: - \: \dfrac {y^{3}}{3} \: = \: 0

Answered by pulakmath007
84

\displaystyle\huge\red{\underline{\underline{Solution}}}

FORMULA TO BE IMPLEMENTED

1. If u & v are two differentiable function then

d(uv) = v \: du + u \: dv

2.

\displaystyle \int\limits_{}^{} x^{n} \, dx =  \frac{ {x}^{n + 1} }{n + 1}  + constant

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

 \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}

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