What is the solution for y^{''}-2y^'+37y=0 ?

The solution for y^{''}-2y^'+37y=0 is y=e^t(c_{1}cos(6t)+c_{2}sin(6t))

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y^{''}-2y^'+37y=0

y^{^{''}} - 2 y^{^{'}} + 37 y = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 37 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))

x \to 0 + lim (\frac{x ^{x} - 1}{x})

\frac{d y}{d x} = sec (x) + y tan (x)

\int \frac{2}{9} x^{- \frac{1}{9}} d x

f (x) = x^{4} - 20 x^{2} + 64

\int (sin^{4} (3 x) cos^{3} (3 x)) d x

What is the solution for y^{''}-2y^'+37y=0 ?
The solution for y^{''}-2y^'+37y=0 is y=e^t(c_{1}cos(6t)+c_{2}sin(6t))