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Popular >

y^'=2y^2+xy^2,y(0)=1

Solution

y^{^{'}} = 2 y^{2} + x y^{2}, y (0) = 1

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

derivative of 1/(5-2x)

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

integral of (9-162x)/(sqrt(64-81x^2))

\int \frac{9 - 162 x}{64 - 81 x ^{2}} d x

integral of-2x*ln(x)

\int - 2 x \cdot ln (x) d x

tangent of x^2-1/x tangent of

x^{2} - \frac{1}{x}

integral of ((ln(x))^7)/x

\int \frac{( ln ( x ) ) ^{7}}{x} d x

Frequently Asked Questions (FAQ)

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