What is the solution for 2y^{''}+18y=5cos(3t) ?

The solution for 2y^{''}+18y=5cos(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(5tsin(3t))/(12)

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2y^{''}+18y=5cos(3t)

2 y^{^{''}} + 18 y = 5 cos (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{5 t sin ( 3 t )}{12}

Solution steps

2 y^{''} (t) + 18 y = 5 cos (3 t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{5 t sin ( 3 t )}{12}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{5 t sin ( 3 t )}{12}

\frac{d}{d x} (\frac{1}{( - x ^{2} + x - 2 ) ^{5}})

x^{2} y^{^{''}} + 5 x y^{^{'}} + 4 y = 0

\frac{d y}{d x} = \frac{1}{x ^{2}} - \frac{2 y}{x}

\frac{x ^{3}}{( x + 1 ) ^{2}}

\frac{d}{d θ} (2 cos (θ) + 2 sin (θ))

What is the solution for 2y^{''}+18y=5cos(3t) ?
The solution for 2y^{''}+18y=5cos(3t) is y=c_{1}cos(3t)+c_{2}sin(3t)+(5tsin(3t))/(12)