What is the solution for y^{''}-5y^'+6y=2sin(4x) ?

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

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y^{''}-5y^'+6y=2sin(4x)

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 sin (4 x)

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = 2 sin (4 x)

Solve linear ODE:

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

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

y^{^{''}} + 7 y^{^{'}} + 10 y = e^{3 x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''''}} - y = 15 t + cos (t)

\frac{d ^{2}}{d t ^{2}} (θ (t)) + 4 θ (t) = csc (2) (2 t)

y^{^{''}} - y^{^{'}} - 2 y = \frac{1}{5}

y^{^{''}} - 9 y = 5 e^{3 x}

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