What is the solution for y^{''}-8y+17y=0 ?

The solution for y^{''}-8y+17y=0 is y=c_{1}cos(3t)+c_{2}sin(3t)

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y^{''}-8y+17y=0

y^{^{''}} - 8 y + 17 y = 0

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Solution steps

y^{''} (t) - 8 y + 17 y = 0

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y^{^{''}} + 14 y^{^{'}} + 48 y = 0, y (0) = 9, y^{^{'}} (0) = - 66

y^{^{''''}} = 625 y

2 y^{^{''}} + 8 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} + 4 y + 5 y = 0

\frac{d ^{2} y}{d x ^{2}} + 10 \frac{d y}{d x} + 41 y = 0

What is the solution for y^{''}-8y+17y=0 ?
The solution for y^{''}-8y+17y=0 is y=c_{1}cos(3t)+c_{2}sin(3t)