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Popular >

y^'+2xy^2=0

Solution

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

Solution

y = - \frac{1}{- x ^{2} + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{- x ^{2} + c _{1}}

y = - \frac{1}{- x ^{2} + c _{1}}

Popular Examples

integral of pi/(1.2-2x)

\int \frac{π}{1.2 - 2 x} d x

derivative of sqrt(18-x^2)-x

\frac{d}{d x} (18 - x^{2} - x)

area f(x)=10x-x^2,g(x)=3x area

f (x) = 10 x - x^{2}, g (x) = 3 x

sum from n=1 to infinity}((4^{n+1 of))/(5^n)

n = 1 \sum \infty \frac{( 4 ^{n + 1} )}{5 ^{n}}

tangent of f(x)= 1/3 x^3-5x^2+26x+23 tangent of

f (x) = \frac{1}{3} x^{3} - 5 x^{2} + 26 x + 23

Frequently Asked Questions (FAQ)

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