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Popular >

y^'+x^{-1}y=xy^2

Solution

y^{^{'}} + x^{- 1} y = x y^{2}

Solution

y = \frac{1}{x ( - x + c _{1} )}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{1}{x ( - x + c _{1} )}

y = \frac{1}{x ( - x + c _{1} )}

Popular Examples

tangent of f(x)=e^x,\at x=0 tangent of

f (x) = e^{x}, at x = 0

f(x)=7x^3+7x^2+7x+3

f (x) = 7 x^{3} + 7 x^{2} + 7 x + 3

derivative of (x^4-1^3(x^3+1)^4)

\frac{d}{d x} ((x^{4} - 1)^{3} (x^{3} + 1)^{4})

integral from 1 to 2 of 1/(8-3t)

\int_{1}^{2} \frac{1}{8 - 3 t} d t

tangent of f(x)=x^2-1,\at x=0 tangent of

f (x) = x^{2} - 1, at x = 0

Frequently Asked Questions (FAQ)

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