What is the solution for y^{''}-3y^'-4y=1-8x^2+48e^{-4x} ?

The solution for y^{''}-3y^'-4y=1-8x^2+48e^{-4x} is y=c_{1}e^{4x}+c_{2}e^{-x}+2x^2-3x+3+2e^{-4x}

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y^{''}-3y^'-4y=1-8x^2+48e^{-4x}

y^{^{''}} - 3 y^{^{'}} - 4 y = 1 - 8 x^{2} + 48 e^{- 4 x}

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

Solution steps

y^{''} (x) - 3 y^{'} (x) - 4 y = 1 - 8 x^{2} + 48 e^{- 4 x}

Solve linear ODE:

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

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

What is the solution for y^{''}-3y^'-4y=1-8x^2+48e^{-4x} ?
The solution for y^{''}-3y^'-4y=1-8x^2+48e^{-4x} is y=c_{1}e^{4x}+c_{2}e^{-x}+2x^2-3x+3+2e^{-4x}