What is the solution for (x+1)y^'+xy=0 ?

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

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(x+1)y^'+xy=0

(x + 1) y^{^{'}} + x y = 0

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

Solution steps

(x + 1) y^{'} (x) + x 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}}

\frac{d}{d x} (\frac{2 x ^{2} - 1}{3 x ^{3} - 1})

\int e^{- x} sin^{2} (x) d x

\frac{d x}{d t} = \frac{4 x}{t}, x (1) = 1

\int \frac{e ^{2 x} ( 1 + 2 x ln ( x ) )}{4 x} d x

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

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