What is the solution for y^'=2+3x+8y+12xy ?

The solution for y^'=2+3x+8y+12xy is y=(e^{8x+6x^2+c_{1}})/4-1/4

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y^'=2+3x+8y+12xy

y^{^{'}} = 2 + 3 x + 8 y + 12 x y

y = \frac{e ^{8 x + 6 x^{2} + c_{1}}}{4} - \frac{1}{4}

Solution steps

y^{'} (x) = 2 + 3 x + 8 y + 12 x y

Solve separable ODE:

y = \frac{e ^{8 x + 6 x^{2} + c_{1}}}{4} - \frac{1}{4}

y = \frac{e ^{8 x + 6 x^{2} + c_{1}}}{4} - \frac{1}{4}

- 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}

\frac{d}{d x} (e^{- 2 x} y)

\frac{d}{d x} (sec^{2} (2 x) - tan^{2} (2 x))

\int \frac{28 x ^{2}}{4 - x ^{2}} d x

\int \frac{x ^{3}}{( x ^{2} + 4 x + 5 ) ^{2}} d x

What is the solution for y^'=2+3x+8y+12xy ?
The solution for y^'=2+3x+8y+12xy is y=(e^{8x+6x^2+c_{1}})/4-1/4