What is the solution for y^{''}-6y^'+14y=0 ?

The solution for y^{''}-6y^'+14y=0 is y=e^{3t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))

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y^{''}-6y^'+14y=0

y^{^{''}} - 6 y^{^{'}} + 14 y = 0

y = e^{3 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 14 y = 0

Solve linear ODE:

y = e^{3 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{3 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y^{^{''}} - y^{^{'}} + 6 y = 0

x^{^{''}} + a^{2} x = 0

y^{^{''}} - 14 y^{^{'}} + 40 y = 0

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 4, y^{^{'}} (0) = 5

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

What is the solution for y^{''}-6y^'+14y=0 ?
The solution for y^{''}-6y^'+14y=0 is y=e^{3t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))