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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

tangent of y=ln(x^2),(5,ln(25))tangent of

y = ln (x^{2}), (5, ln (25))

integral of 1/(x(\sqrt[7]{x))^1}

derivative of f(x)=3x^8 derivative of

f (x) = 3 x^{8}

limit as x approaches 0 of x^{sin(2x)}

x \to 0 lim (x^{s i n (2 x)})

(\partial)/(\partial x)((4x)/(y^3))

\frac{\partial}{\partial x} (\frac{4 x}{y ^{3}})

Frequently Asked Questions (FAQ)

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