What is the solution for y^{''}-5y^'+6y=8e^{4x} ?

The solution for y^{''}-5y^'+6y=8e^{4x} is y=c_{1}e^{3x}+c_{2}e^{2x}+4e^{4x}

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y^{''}-5y^'+6y=8e^{4x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 8 e^{4 x}

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = 8 e^{4 x}

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = 20 cos (x)

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 50

D^{2} (D - 1) y = 8 e^{(2 x)}

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 e^{2 t}

y^{^{''}} + 16 y = 13 sec^{2} (4 t)

What is the solution for y^{''}-5y^'+6y=8e^{4x} ?
The solution for y^{''}-5y^'+6y=8e^{4x} is y=c_{1}e^{3x}+c_{2}e^{2x}+4e^{4x}