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If the solution of the differential equation  \(\frac{dy}{dx}=\frac{1}{x, \cos y+\sin 2y}\;\; is\;\; x=ce^{\sin y}-k(1+\sin y)), then k =

Option: 1 1

Option: 2 2

Option: 3 3

Option: 4 4

Answers (1)

best_answer

 

Linear Differential Equation -

 

\frac{dy}{dx}+Py= Q

yhh- wherein

P, Q are functions of x alone.

 

 

Linear Differential Equation -

Multiply by e^{SPdx}  which is the Integrating factor

- wherein

P is the function of x alone

 

 

\frac{dy}{dx}=\frac{1}{x\cos y+2\sin y \cos y}

\therefore \frac{dx}{dy}=x\cos y+2\sin y \cos y

\Rightarrow \frac{dx}{dy}+(-\cos y )x =2\sin y \cos y

\therefore I.F.=e^{-\int \cos y dy}=e^{-\sin y}

\therefore The\:\:solution\:\:is

x.e^{-\sin y}=2\int e^{-\sin y}. \sin y \cos y \:\:dy

\Rightarrow x.e^{-\sin y}= -2 \sin y\: e ^{-\sin y}-2 \int (-e^{-\sin y }) \cos y \:\:dy

\Rightarrow x.e^{-\sin y}= -2 \sin y\: e ^{-\sin y}+2 \int (e^{-\sin y }) \cos y \:\:dy

                        =-2 \sin y\:\: e ^{- \sin y}-2\: e ^{-\sin y }+c

\Rightarrow x=-2 \sin y -2 + c\:\: e ^{\sin y}=c\:\:e^{\sin y}-2(1+\sin y)

\therefore k=2

 

 

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Shailly goel

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