What is the solution for y^{''}-2y^'+17y=0 ?

The solution for y^{''}-2y^'+17y=0 is y=e^t(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}-2y^'+17y=0

y^{^{''}} - 2 y^{^{'}} + 17 y = 0

y = e^{t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) - 2 y^{'} (t) + 17 y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{t} (c_{1} cos (4 t) + c_{2} sin (4 t))

x \to 1 lim (\frac{- 1}{x})

x \to 0 + lim ((x + 1)^{I n (x)})

\frac{d}{d x} (- 5 sin (x) (x + 3 sin (x)))

(x^{9} - 2)^{3}

x \to 1 lim (2 - x)

What is the solution for y^{''}-2y^'+17y=0 ?
The solution for y^{''}-2y^'+17y=0 is y=e^t(c_{1}cos(4t)+c_{2}sin(4t))