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Popular >

y^'=9y^2+xy^2,y(0)=1

Solution

y^{^{'}} = 9 y^{2} + x y^{2}, y (0) = 1

Solution

y = - \frac{2}{18 x + x ^{2} - 2}

Solution steps

y^{'} (x) = 9 y^{2} + x y^{2}

Solve separable ODE:

y = - \frac{2}{18 x + x ^{2} - 2}

y = - \frac{2}{18 x + x ^{2} - 2}

Popular Examples

(\partial)/(\partial x)((41-x^2-y^2)^{1/2})

\frac{\partial}{\partial x} ((41 - x^{2} - y^{2})^{\frac{1}{2}})

integral from 0 to 1 of xe^{-4x}

\int_{0}^{1} x e^{- 4 x} d x

integral of 1/(xsqrt(25x^2+16))

\int \frac{1}{x 25 x ^{2} + 16} d x

sqrt(x)+sqrt(y)(dy)/(dx)=0

x + y \frac{d y}{d x} = 0

derivative of 3e^{2x}+2e^{2x}x

\frac{d}{d x} (3 e^{2 x} + 2 e^{2 x} x)

Frequently Asked Questions (FAQ)

What is the solution for y^'=9y^2+xy^2,y(0)=1 ?
The solution for y^'=9y^2+xy^2,y(0)=1 is y=-2/(18x+x^2-2)