What is the solution for (dy)/(dx)=yx^2+yx,y(0)=-1 ?

The solution for (dy)/(dx)=yx^2+yx,y(0)=-1 is y=-1/(e^{\frac{-2x^3-3x^2){6}}}

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(dy)/(dx)=yx^2+yx,y(0)=-1

\frac{d y}{d x} = y x^{2} + y x, y (0) = - 1

y = - \frac{1}{e ^{\frac{- 2 x ^{3} - 3 x ^{2}}{6}}}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{e ^{\frac{- 2 x ^{3} - 3 x ^{2}}{6}}}

y = - \frac{1}{e ^{\frac{- 2 x ^{3} - 3 x ^{2}}{6}}}

\int x^{2} 2^{x} d x

\int_{1}^{7} - \frac{1}{3} y^{2} + 3 y d y

x \frac{d y}{d x} + x y = 1 - y

n = 0 \sum \infty \frac{x ^{n}}{3 ^{n}}

x^{2} y y^{^{'}} + 3 x y^{2} = 0

What is the solution for (dy)/(dx)=yx^2+yx,y(0)=-1 ?
The solution for (dy)/(dx)=yx^2+yx,y(0)=-1 is y=-1/(e^{\frac{-2x^3-3x^2){6}}}