solution of diffrential equation dy/dx = (x+y) \cot x+ \csc x -1   is 

  • Option 1)

    x+y - \cos x =c \sin x

  • Option 2)

    x+y +\cos x =c \sin x

  • Option 3)

    x-y - \cos x =c \sin x

  • Option 4)

    x+y + \cos x =-c \sin x

 

Answers (1)
G gaurav

As we have learned

Extended Form of linear Differential Equation -

Sometimes, a differential equation is not linear but it can be converted into a linear differential equation

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\frac{dy}{dx}=(x+y )\cot x + \csc x -1

\frac{dy}{dx}+1= (x+y )\cot x + \csc x

Let us put x+y = t \Rightarrow 1+ \frac{dy}{dx}= \frac{dt }{dx}

\Rightarrow \frac{dt }{dx}= (\cot x )t + \csc x \Rightarrow \frac{dt}{dx} + (-\cot x) t = \csc x

This is comparable with , dy/dx + py = Q.. so.. p = - \cot x

I.F = e^{-\cot x dx } = \csc x

multiplying both sides by integrating factor , we get \rightarrow 

\csc \frac{dt}{dx}- (\csc x \cot x )t = \csc^2x

\Rightarrow \frac{d}{dx}(\csc x \cdot t ) = \csc^2 x

\Rightarrow\int {d}(\csc x \cdot t ) = \int \csc^2 x dx

\Rightarrow t \csc x + \cot x= C

(x+y )+ \cos x = C \sin x

 

 

 


Option 1)

x+y - \cos x =c \sin x

Option 2)

x+y +\cos x =c \sin x

Option 3)

x-y - \cos x =c \sin x

Option 4)

x+y + \cos x =-c \sin x

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