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

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

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2y^{''}+18y+5=0,y(0)=0,y^'(0)=0

2 y^{^{''}} + 18 y + 5 = 0, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{5}{18} cos (3 t) - \frac{5}{18}

Solution steps

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

Solve linear ODE:

y = \frac{5}{18} cos (3 t) - \frac{5}{18}

y = \frac{5}{18} cos (3 t) - \frac{5}{18}

y, (d^{3} + 4 d) y = 2 t + t^{2}

y^{^{''}} - 9 y = x

y^{^{''}} + y = 2 \cdot e^{x}

y^{^{''}} - y^{^{'}} - 6 y = 10 x e^{3 x} + 17 e^{3 x}

y^{^{''}} - 4 y = x + sin (x)

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