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Popular >

y^'+(x+2)y=0

Solution

y^{^{'}} + (x + 2) y = 0

Solution

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

Solution steps

y^{'} (x) + (x + 2) y = 0

Solve separable ODE:

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

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

Popular Examples

(e^y+1)^2e^{-y}dx+(e^x+1)^7e^{-x}dy=0

(e^{y} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{7} e^{- x} d y = 0

e^x(dy)/(dx)+y^2=xy^2

e^{x} \frac{d y}{d x} + y^{2} = x y^{2}

y \cdot d x = - (1 - x) d y

(dy)/(dt)=4yt^2

\frac{d y}{d t} = 4 y t^{2}

(e^y+21)^2e^{-y}dx+(e^x+1)^3e^{-x}dy=0

(e^{y} + 21)^{2} e^{- y} d x + (e^{x} + 1)^{3} e^{- x} d y = 0

Frequently Asked Questions (FAQ)

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