What is the solution for y^{''}-5y^'+6y=4e^{4x} ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 3 y = 5 sin (4 t), y (0) = 2, y^{^{'}} (0) = 1

y^{^{''}} - 3 y^{^{'}} - 4 y = - 16 t, y (0) = - 4, y^{^{'}} (0) = - 5

y^{^{''}} + 4 y = 5 sin (2 x) - 7 cos (2 x)

y^{^{''}} - 4 y^{^{'}} = 6 e^{3 x} - 3 e^{- x}

2 y^{^{''}} + 8 y^{^{'}} - y = 16

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