What is the solution for 4y^{''}-y^'+17y=0 ?

The solution for 4y^{''}-y^'+17y=0 is y=e^{t/8}(c_{1}cos((sqrt(271)t)/8)+c_{2}sin((sqrt(271)t)/8))

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4y^{''}-y^'+17y=0

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

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

Solution steps

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

Solve linear ODE:

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

y = e^{\frac{t}{8}} (c_{1} cos (\frac{271 t}{8}) + c_{2} sin (\frac{271 t}{8}))

y^{^{''}} + \frac{4}{3} y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{10}{3}

a (x) y^{^{''}} + b (x) y^{^{'}} + c (x) y = 0

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} + 7 y = 0

y^{^{''}} + 15 y^{^{'}} = 0, y (0) = - 27, y^{^{'}} (0) = 15

u^{^{''}} - H (t) = 0

What is the solution for 4y^{''}-y^'+17y=0 ?
The solution for 4y^{''}-y^'+17y=0 is y=e^{t/8}(c_{1}cos((sqrt(271)t)/8)+c_{2}sin((sqrt(271)t)/8))