What is the solution for y^{''}-9y^'+18y=-4cos(3t) ?

The solution for y^{''}-9y^'+18y=-4cos(3t) is y=c_{1}e^{6t}+c_{2}e^{3t}+2/15 sin(3t)-2/45 cos(3t)

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y^{''}-9y^'+18y=-4cos(3t)

y^{^{''}} - 9 y^{^{'}} + 18 y = - 4 cos (3 t)

y = c_{1} e^{6 t} + c_{2} e^{3 t} + \frac{2}{15} sin (3 t) - \frac{2}{45} cos (3 t)

Solution steps

y^{''} (t) - 9 y^{'} (t) + 18 y = - 4 cos (3 t)

Solve linear ODE:

y = c_{1} e^{6 t} + c_{2} e^{3 t} + \frac{2}{15} sin (3 t) - \frac{2}{45} cos (3 t)

y = c_{1} e^{6 t} + c_{2} e^{3 t} + \frac{2}{15} sin (3 t) - \frac{2}{45} cos (3 t)

f (x) = \frac{14}{1 + e ^{- x}}

\int e^{x} 5 sin (2 x) d x

x \to 9 lim (\frac{x ^{2} - 17 x + 72}{x - 9})

(t - 1) (t - 1)

9 e^{x} cos (x)

What is the solution for y^{''}-9y^'+18y=-4cos(3t) ?
The solution for y^{''}-9y^'+18y=-4cos(3t) is y=c_{1}e^{6t}+c_{2}e^{3t}+2/15 sin(3t)-2/45 cos(3t)