# 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) $x+y + \cos x =-c \sin x$

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

-

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

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

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)

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

Exams
Articles
Questions