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

The solution for y^{''}-5y^'+6y=sin(4x) is y=c_{1}e^{3x}+c_{2}e^{2x}-1/50 sin(4x)+1/25 cos(4x)

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

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

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{1}{50} sin (4 x) + \frac{1}{25} cos (4 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{1}{50} sin (4 x) + \frac{1}{25} cos (4 x)

y = c_{1} e^{3 x} + c_{2} e^{2 x} - \frac{1}{50} sin (4 x) + \frac{1}{25} cos (4 x)

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

y^{^{''}} - 7 y^{^{'}} + 10 y = x^{2} - 4 + e^{x}

y^{^{''}} + 8 y^{^{'}} + 16 y = t^{- 2} e^{- 4 t}

y^{^{''}} + 9 y = 10 e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{'''}} - y^{^{'}} = 5 sin (t)

What is the solution for y^{''}-5y^'+6y=sin(4x) ?
The solution for y^{''}-5y^'+6y=sin(4x) is y=c_{1}e^{3x}+c_{2}e^{2x}-1/50 sin(4x)+1/25 cos(4x)