What is the solution for y^{''}-18y^'+162y=0 ?

The solution for y^{''}-18y^'+162y=0 is y=e^{9t}(c_{1}cos(9t)+c_{2}sin(9t))

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y^{''}-18y^'+162y=0

y^{^{''}} - 18 y^{^{'}} + 162 y = 0

y = e^{9 t} (c_{1} cos (9 t) + c_{2} sin (9 t))

Solution steps

y^{''} (t) - 18 y^{'} (t) + 162 y = 0

Solve linear ODE:

y = e^{9 t} (c_{1} cos (9 t) + c_{2} sin (9 t))

y = e^{9 t} (c_{1} cos (9 t) + c_{2} sin (9 t))

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 17 y = 0

x^{^{'''}} + 9 x^{^{''}} + 15 x^{^{'}} - 25 x = 0

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

a y^{^{''}} + b y^{^{'}} + cy = 0, y (0) = 0

y^{^{''}} - 10 y^{^{'}} + 41 y = 0, y (0) = 10, y^{^{'}} (0) = 7

What is the solution for y^{''}-18y^'+162y=0 ?
The solution for y^{''}-18y^'+162y=0 is y=e^{9t}(c_{1}cos(9t)+c_{2}sin(9t))