What is the solution for 2y^{''}+18y=0,y(0)=2,y^'(0)=15 ?

The solution for 2y^{''}+18y=0,y(0)=2,y^'(0)=15 is y=2cos(3t)+5sin(3t)

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2y^{''}+18y=0,y(0)=2,y^'(0)=15

2 y^{^{''}} + 18 y = 0, y (0) = 2, y^{^{'}} (0) = 15

y = 2 cos (3 t) + 5 sin (3 t)

Solution steps

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

Solve linear ODE:

y = 2 cos (3 t) + 5 sin (3 t)

y = 2 cos (3 t) + 5 sin (3 t)

y^{^{'''''}} + 20 y^{^{'''}} = 0

\frac{5}{8} x^{^{''}} + 40 x = 0, x (0) = \frac{1}{3}, x^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = a, y^{^{'}} (0) = 10

y^{^{''}} + 2 α y^{^{'}} + α^{2} y = 0

y^{^{''}} - y^{^{'}} - 2 \cdot y = 0

What is the solution for 2y^{''}+18y=0,y(0)=2,y^'(0)=15 ?
The solution for 2y^{''}+18y=0,y(0)=2,y^'(0)=15 is y=2cos(3t)+5sin(3t)