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Popular >

y^'=4y^2+xy^2,y(0)=1

Solution

y^{^{'}} = 4 y^{2} + x y^{2}, y (0) = 1

Solution

y = - \frac{2}{8 x + x ^{2} - 2}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{8 x + x ^{2} - 2}

y = - \frac{2}{8 x + x ^{2} - 2}

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Frequently Asked Questions (FAQ)

What is the solution for y^'=4y^2+xy^2,y(0)=1 ?
The solution for y^'=4y^2+xy^2,y(0)=1 is y=-2/(8x+x^2-2)