What is the solution for 2y^{''}+18y+5=0 ?

The solution for 2y^{''}+18y+5=0 is y=c_{1}cos(3t)+c_{2}sin(3t)-5/18

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2y^{''}+18y+5=0

2 y^{^{''}} + 18 y + 5 = 0

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

Solution steps

2 y^{''} (t) + 18 y + 5 = 0

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} - 6 y = 2

y^{^{''}} + 2 y^{^{'}} - 3 y = 5 sin (3 x)

y^{^{''}} - y^{^{'}} - 1 = 0

y^{^{'''}} + 3 y^{^{''}} - 4 y^{^{'}} - 12 y = 10 e^{3 x}

\frac{d ^{2} y}{d x ^{2}} + 9 y = tan (3 x)

What is the solution for 2y^{''}+18y+5=0 ?
The solution for 2y^{''}+18y+5=0 is y=c_{1}cos(3t)+c_{2}sin(3t)-5/18