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Popular >

y^'=7y^2+xy^2,y(0)=1

Solution

y^{^{'}} = 7 y^{2} + x y^{2}, y (0) = 1

Solution

y = - \frac{2}{14 x + x ^{2} - 2}

Solution steps

y^{'} (x) = 7 y^{2} + x y^{2}

Solve separable ODE:

y = - \frac{2}{14 x + x ^{2} - 2}

y = - \frac{2}{14 x + x ^{2} - 2}

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\int \frac{1}{x ^{3} + 27} d x

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x \to 0 lim (\frac{x - 4}{x - 4})

integral of sin(2x)(e^{cos(x)})

\int sin (2 x) (e^{c o s (x)}) d x

integral of arccosh(x)

\int arccosh (x) d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=7y^2+xy^2,y(0)=1 ?
The solution for y^'=7y^2+xy^2,y(0)=1 is y=-2/(14x+x^2-2)