What is the solution for xydx-(x+2)dy=0 ?

The solution for xydx-(x+2)dy=0 is y=(e^{x+2+c_{1}})/(x^2+4x+4)

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xydx-(x+2)dy=0

x y d x - (x + 2) d y = 0

y = \frac{e ^{x + 2 + c_{1}}}{x ^{2} + 4 x + 4}

Solution steps

x y d x - (x + 2) d y = 0

Solve separable ODE:

y = \frac{e ^{x + 2 + c_{1}}}{x ^{2} + 4 x + 4}

y = \frac{e ^{x + 2 + c_{1}}}{x ^{2} + 4 x + 4}

\int \frac{7}{x ^{4}} d x

f (x) = (1 + x)^{- 4}

\frac{x}{( \frac{x}{10} ) ^{^{'}}}

x^{2} - y^{2} = 64, (8, 0)

2 x - x^{2}, 8 - 4 x, 4 x

What is the solution for xydx-(x+2)dy=0 ?
The solution for xydx-(x+2)dy=0 is y=(e^{x+2+c_{1}})/(x^2+4x+4)