What is the solution for y^{''}-y=6e^{2x},y(0)=4,y^'(0)=4 ?

The solution for y^{''}-y=6e^{2x},y(0)=4,y^'(0)=4 is y=e^x+e^{-x}+2e^{2x}

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y^{''}-y=6e^{2x},y(0)=4,y^'(0)=4

y^{^{''}} - y = 6 e^{2 x}, y (0) = 4, y^{^{'}} (0) = 4

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

Solution steps

y^{''} (x) - y = 6 e^{2 x}

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} cos (x)

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{2 t}

(D^{4} - 1) \cdot y = sin (2 x)

y^{^{''}} + y^{^{'}} - 7 y = 42 t

y^{^{''}} + y = 2, y (0) = 2, y (0) = 0

What is the solution for y^{''}-y=6e^{2x},y(0)=4,y^'(0)=4 ?
The solution for y^{''}-y=6e^{2x},y(0)=4,y^'(0)=4 is y=e^x+e^{-x}+2e^{2x}