What is the solution for y^{''}-y^'-6y=51cos(3x) ?

The solution for y^{''}-y^'-6y=51cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-17/26 sin(3x)-85/26 cos(3x)

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y^{''}-y^'-6y=51cos(3x)

y^{^{''}} - y^{^{'}} - 6 y = 51 cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{17}{26} sin (3 x) - \frac{85}{26} cos (3 x)

Solution steps

y^{''} (x) - y^{'} (x) - 6 y = 51 cos (3 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{17}{26} sin (3 x) - \frac{85}{26} cos (3 x)

y = c_{1} e^{3 x} + c_{2} e^{- 2 x} - \frac{17}{26} sin (3 x) - \frac{85}{26} cos (3 x)

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

y^{^{''}} - a \cdot y + b = 0

y^{^{''}} + 16 y^{^{'}} + 100 y = 10, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = cos (2 x)

y^{^{''}} + π^{2} y = \frac{5}{ln ( 10 )} sin (x ln (2))

What is the solution for y^{''}-y^'-6y=51cos(3x) ?
The solution for y^{''}-y^'-6y=51cos(3x) is y=c_{1}e^{3x}+c_{2}e^{-2x}-17/26 sin(3x)-85/26 cos(3x)