What is the solution for 2(d^2x)/(dt^2)+18x+5=0 ?

The solution for 2(d^2x)/(dt^2)+18x+5=0 is x=c_{1}cos(3t)+c_{2}sin(3t)-5/18

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2(d^2x)/(dt^2)+18x+5=0

2 \frac{d ^{2} x}{d t ^{2}} + 18 x + 5 = 0

x = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{5}{18}

Solution steps

2 \frac{d ^{2} x}{d t ^{2}} + 18 x + 5 = 0

Solve linear ODE:

x = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{5}{18}

x = c_{1} cos (3 t) + c_{2} sin (3 t) - \frac{5}{18}

y^{^{''}} + 9 y = t

y^{^{''}} + 2 y^{^{'}} + y = 3 x^{4} e^{- x}

9 y^{^{''}} - 6 y^{^{'}} + y = 9 - x^{2}

y^{^{''}} - y^{^{'}} - 2 y = sin (3 t), y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} - 2 y^{^{'}} + 5 y = 1 + t, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for 2(d^2x)/(dt^2)+18x+5=0 ?
The solution for 2(d^2x)/(dt^2)+18x+5=0 is x=c_{1}cos(3t)+c_{2}sin(3t)-5/18