What is the solution for y^{''}-4y=12x,y(0)=4,y^'(0)=1 ?

The solution for y^{''}-4y=12x,y(0)=4,y^'(0)=1 is y=3e^{2x}+e^{-2x}-3x

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y^{''}-4y=12x,y(0)=4,y^'(0)=1

y^{^{''}} - 4 y = 12 x, y (0) = 4, y^{^{'}} (0) = 1

y = 3 e^{2 x} + e^{- 2 x} - 3 x

Solution steps

y^{''} (x) - 4 y = 12 x

Solve linear ODE:

y = 3 e^{2 x} + e^{- 2 x} - 3 x

y = 3 e^{2 x} + e^{- 2 x} - 3 x

y^{^{''}} - 6 y^{^{'}} + 9 y = x

y^{^{''}} - 2 y^{^{'}} - 15 y = 384 e^{- t}

y^{^{''}} + 4 y = - 1, y (\frac{π}{8}) = \frac{3}{4}, y^{^{'}} (\frac{π}{8}) = 2

y^{^{''}} + 5 y^{^{'}} + 4 y = 12

y^{^{''}} - 22 y^{^{'}} + 121 y = \frac{e ^{- 11 x}}{1 + x ^{2}}

What is the solution for y^{''}-4y=12x,y(0)=4,y^'(0)=1 ?
The solution for y^{''}-4y=12x,y(0)=4,y^'(0)=1 is y=3e^{2x}+e^{-2x}-3x