What is the solution for 1y^{''}+0y^'+9y=0,y(0)=6,y^'(0)=-17 ?

The solution for 1y^{''}+0y^'+9y=0,y(0)=6,y^'(0)=-17 is y=6cos(3t)-17/3 sin(3t)

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1y^{''}+0y^'+9y=0,y(0)=6,y^'(0)=-17

1 y^{^{''}} + 0 y^{^{'}} + 9 y = 0, y (0) = 6, y^{^{'}} (0) = - 17

y = 6 cos (3 t) - \frac{17}{3} sin (3 t)

Solution steps

1 y^{''} (t) + 0 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = 6 cos (3 t) - \frac{17}{3} sin (3 t)

y = 6 cos (3 t) - \frac{17}{3} sin (3 t)

\frac{\frac{d ^{2} f}{d x ^{2}}}{f ( x )} = 1

y^{^{''}} + y^{^{'}} + 49 y = 0

4 y^{^{''}} - 20 y^{^{'}} - 169 y = 0

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 15 y = 0

y^{^{''}} + \frac{a}{b} y^{^{'}} = 0

What is the solution for 1y^{''}+0y^'+9y=0,y(0)=6,y^'(0)=-17 ?
The solution for 1y^{''}+0y^'+9y=0,y(0)=6,y^'(0)=-17 is y=6cos(3t)-17/3 sin(3t)