What is the solution for y^{''}+18y=0,y(0)=-3,y^'(0)=-1 ?

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

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y^{''}+18y=0,y(0)=-3,y^'(0)=-1

y^{^{''}} + 18 y = 0, y (0) = - 3, y^{^{'}} (0) = - 1

y = - 3 cos (32 t) - \frac{1}{3 2} sin (32 t)

Solution steps

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

Solve linear ODE:

y = - 3 cos (32 t) - \frac{1}{3 2} sin (32 t)

y = - 3 cos (32 t) - \frac{1}{3 2} sin (32 t)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 6 y = 0

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 4

y^{^{''}} + 8 y^{^{'}} + 16 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 2

x^{^{''}} + 10 x^{^{'}} + 16 x = 0, x (0) = 1, x^{^{'}} (0) = - 13

4 y^{^{''}} + 8 y^{^{'}} - 21 y = 0

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