What is the solution for -12x^2y+y^'=-8x^2 ?

The solution for -12x^2y+y^'=-8x^2 is y=(e^{4x^3+c_{1}})/3+2/3

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-12x^2y+y^'=-8x^2

- 12 x^{2} y + y^{^{'}} = - 8 x^{2}

y = \frac{e ^{4 x^{3} + c_{1}}}{3} + \frac{2}{3}

Solution steps

- 12 x^{2} y + y^{'} (x) = - 8 x^{2}

Solve separable ODE:

y = \frac{e ^{4 x^{3} + c_{1}}}{3} + \frac{2}{3}

y = \frac{e ^{4 x^{3} + c_{1}}}{3} + \frac{2}{3}

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

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

\frac{d}{d x} (r (θ, x) cos (θ))

\int (2 x + 3)^{2} d x

sin (u)

What is the solution for -12x^2y+y^'=-8x^2 ?
The solution for -12x^2y+y^'=-8x^2 is y=(e^{4x^3+c_{1}})/3+2/3