What is the solution for y^'-2y=3x^2 ?

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

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y^'-2y=3x^2

y^{^{'}} - 2 y = 3 x^{2}

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

Solution steps

y^{'} (x) - 2 y = 3 x^{2}

Solve linear ODE:

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

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

x \to 4 lim (x^{2} - 16)

x \to - 2 lim (\frac{5 x + 3}{2 x - 9})

(2 x^{5} - 3)^{8}

\int_{0}^{2} \frac{( x + 2 )}{( x - 4 )} d x

\frac{\partial}{\partial x} (36 - 6 x^{2} - y^{2})

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