What is the (e^{-y}+1)sin(x)dx=(1+cos(x))dy,y(0)=0 ?

The (e^{-y}+1)sin(x)dx=(1+cos(x))dy,y(0)=0 is 0

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(e^{-y}+1)sin(x)dx=(1+cos(x))dy,y(0)=0

(e^{- y} + 1) sin (x) d x = (1 + cos (x)) d y, y (0) = 0

0

Solution steps

Find

(e^{- y} + 1) sin (x) d x \circ y (0)

given

(e^{- y} + 1) sin (x) d x (x, y) = (1 + cos (x)) d y, y (0) = 0

Expand

(e^{- y} + 1) sin (x) d x \circ y \cdot 0 : 0

= 0

\int p (p + 1)^{5} d p

\frac{\partial}{\partial y} (y - \frac{1}{x ^{2}})

\int tan (x) sec (x) d x

\frac{d y}{d t} = e^{- t}

\frac{\partial}{\partial y} (2 x z)

What is the (e^{-y}+1)sin(x)dx=(1+cos(x))dy,y(0)=0 ?
The (e^{-y}+1)sin(x)dx=(1+cos(x))dy,y(0)=0 is 0