What is the solution for xydx+(x+1)dy=0 ?

The solution for xydx+(x+1)dy=0 is y=e^{-x-1+c_{1}}x+e^{-x-1+c_{1}}

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xydx+(x+1)dy=0

x y d x + (x + 1) d y = 0

y = e^{- x - 1 + c_{1}} x + e^{- x - 1 + c_{1}}

Solution steps

x y d x + (x + 1) d y = 0

Solve separable ODE:

y = e^{- x - 1 + c_{1}} x + e^{- x - 1 + c_{1}}

y = e^{- x - 1 + c_{1}} x + e^{- x - 1 + c_{1}}

\int (x^{2} - 4)^{3} (2 x) d x

f (x) = 7 x - 4 x, at x = 1

g (x) = 9 x e^{5 x}

\int - 5 x^{3} + \frac{3}{4} x^{2} d x

\frac{d}{d x} ((e^{- x} - e^{5}) (e^{x} + e^{- 2}))

What is the solution for xydx+(x+1)dy=0 ?
The solution for xydx+(x+1)dy=0 is y=e^{-x-1+c_{1}}x+e^{-x-1+c_{1}}