What is the solution for y^{''}-5y^'+6y=4e^{4x},y(0)=0,y^'(0)=1 ?

The solution for y^{''}-5y^'+6y=4e^{4x},y(0)=0,y^'(0)=1 is y=-3e^{3x}+e^{2x}+2e^{4x}

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y^{''}-5y^'+6y=4e^{4x},y(0)=0,y^'(0)=1

y^{^{''}} - 5 y^{^{'}} + 6 y = 4 e^{4 x}, y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = 4 e^{4 x}

Solve linear ODE:

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

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

(d^{2} + 4) y = cos (2 x)

y^{^{''}} - 14 y^{^{'}} + 49 y = 21 x + 2

y^{^{''}} + 4 y^{^{'}} + 5 y = - 15 x + 3 e^{- x}

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 2 + e^{- t}

What is the solution for y^{''}-5y^'+6y=4e^{4x},y(0)=0,y^'(0)=1 ?
The solution for y^{''}-5y^'+6y=4e^{4x},y(0)=0,y^'(0)=1 is y=-3e^{3x}+e^{2x}+2e^{4x}