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Popular >

y^'+2xy-x=0

Solution

y^{^{'}} + 2 x y - x = 0

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

\frac{d y}{d t} = 2 c - y

((dy)/(dx))+y=yx(e^{(x+2)})

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

(dy)/(dx)=(e^{7x+y})/(y-4)

\frac{d y}{d x} = \frac{e ^{7 x + y}}{y - 4}

(dP)/(dt)= 1/20 (1-P/(1000))

\frac{d P}{d t} = \frac{1}{20} (1 - \frac{P}{1000})

(x+1)yy^'=y^2+1,y(1)=sqrt(3)

(x + 1) y y^{^{'}} = y^{2} + 1, y (1) = 3

Frequently Asked Questions (FAQ)

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