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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

(dN)/(dt)=N(1-0.0005N)

\frac{d N}{d t} = N (1 - 0.0005 N)

x(dy)/(dx)+y=(xy)^{3/2}

x \frac{d y}{d x} + y = (x y)^{\frac{3}{2}}

derivative of sqrt(y-x)n(y+x)

\frac{d}{d x} (y - x n (y + x))

(\partial)/(\partial y)(x-y-1)

\frac{\partial}{\partial y} (x - y - 1)

limit as x approaches 0 of e^{-x}cos(x)

x \to 0 lim (e^{- x} cos (x))

Frequently Asked Questions (FAQ)

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