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Popular >

y^'+2y=xy^2

Solution

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

Solution

y = \frac{4}{2 x + 1 + 4 c _{1} e ^{2 x}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{4}{2 x + 1 + 4 c _{1} e ^{2 x}}

y = \frac{4}{2 x + 1 + 4 c _{1} e ^{2 x}}

Popular Examples

derivative of f(x)=(2x-8)^4(x^2+x+1)^5 derivative of

f (x) = (2 x - 8)^{4} (x^{2} + x + 1)^{5}

area y=x^3-10x^2+21x,y=-x^3+10x^2-21x area

y = x^{3} - 10 x^{2} + 21 x, y = - x^{3} + 10 x^{2} - 21 x

derivative of e^{3/x}derivative of

e^{\frac{3}{x}}

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

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

(dy)/(dx)+(xy)/(x^2+1)= x/(x^2+1)

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

Frequently Asked Questions (FAQ)

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