What is the solution for y^{''}-2y^'+7y=0 ?

The solution for y^{''}-2y^'+7y=0 is y=e^t(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))

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y^{''}-2y^'+7y=0

y^{^{''}} - 2 y^{^{'}} + 7 y = 0

y = e^{t} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{t} (c_{1} cos (6 t) + c_{2} sin (6 t))

11 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

\frac{d}{d x} (\frac{1}{10 x})

\int_{0}^{l n (2)} 10 e^{2 t} d t

\int (- \frac{1}{x} + 7 e^{x}) d x

\frac{d}{d x} (6 + x^{2})

What is the solution for y^{''}-2y^'+7y=0 ?
The solution for y^{''}-2y^'+7y=0 is y=e^t(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))