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Popular >

y^'+4xy^2=0

Solution

y^{^{'}} + 4 x y^{2} = 0

Solution

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

Solution steps

y^{'} (x) + 4 x y^{2} = 0

Solve separable ODE:

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

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

Popular Examples

(dy)/(1+4y^2)=-(xdx)/(1+x^4)

\frac{d y}{1 + 4 y ^{2}} = - \frac{x d x}{1 + x ^{4}}

integral of (5x-1)/((x+1)(x-2))

\int \frac{5 x - 1}{( x + 1 ) ( x - 2 )} d x

tangent of x^9,\at x=1 tangent of

x^{9}, at x = 1

integral of 4/((x-1)(x+3))

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

tangent of f(x)=x^2sqrt(x^3+1),\at x=5 tangent of

f (x) = x^{2} x^{3} + 1, at x = 5

Frequently Asked Questions (FAQ)

What is the solution for y^'+4xy^2=0 ?
The solution for y^'+4xy^2=0 is y=-1/(-2x^2+c_{1)}