What is the solution for y^{''}-2y^'-3y=5e^{4x},y(0)=1,y^'(0)=0 ?

The solution for y^{''}-2y^'-3y=5e^{4x},y(0)=1,y^'(0)=0 is y=-e^{3x}+e^{-x}+e^{4x}

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y^{''}-2y^'-3y=5e^{4x},y(0)=1,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} - 3 y = 5 e^{4 x}, y (0) = 1, y^{^{'}} (0) = 0

y = - e^{3 x} + e^{- x} + e^{4 x}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 5 e^{4 x}

Solve linear ODE:

y = - e^{3 x} + e^{- x} + e^{4 x}

y = - e^{3 x} + e^{- x} + e^{4 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 6

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

y^{^{''}} - 2 y^{^{'}} + 5 y = sin (x)

y^{^{''}} + 4 y = 4 sec^{2} (2 t)

y^{^{''}} - 2 y^{^{'}} + y = 10 e^{- 2 x} cos (x)

What is the solution for y^{''}-2y^'-3y=5e^{4x},y(0)=1,y^'(0)=0 ?
The solution for y^{''}-2y^'-3y=5e^{4x},y(0)=1,y^'(0)=0 is y=-e^{3x}+e^{-x}+e^{4x}