What is the solution for -2x^2y+y^'=-7x^2 ?

The solution for -2x^2y+y^'=-7x^2 is y=(e^{(2x^3+c_{1})/3})/(2)+7/2

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-2x^2y+y^'=-7x^2

- 2 x^{2} y + y^{^{'}} = - 7 x^{2}

y = \frac{e ^{\frac{2 x ^{3} + c _{1}}{3}}}{2} + \frac{7}{2}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{\frac{2 x ^{3} + c _{1}}{3}}}{2} + \frac{7}{2}

y = \frac{e ^{\frac{2 x ^{3} + c _{1}}{3}}}{2} + \frac{7}{2}

x \cdot (3 \cdot x - 4)^{0.5}

\frac{\partial}{\partial y} (x z - 5 x^{2} y^{3} z^{4})

\frac{d}{d x} (\frac{- 10 x + 4}{3 x - 8})

y = x^{2} + x

\int (x + y) d x

What is the solution for -2x^2y+y^'=-7x^2 ?
The solution for -2x^2y+y^'=-7x^2 is y=(e^{(2x^3+c_{1})/3})/(2)+7/2