What is the solution for 4y^{''}-16y^'+17y=0 ?

The solution for 4y^{''}-16y^'+17y=0 is y=e^{2t}(c_{1}cos(t/2)+c_{2}sin(t/2))

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4y^{''}-16y^'+17y=0

4 y^{^{''}} - 16 y^{^{'}} + 17 y = 0

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

Solution steps

4 y^{''} (t) - 16 y^{'} (t) + 17 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 0, y (0) = 3

y^{^{''}} + (a - 1) y^{^{'}} + b y = 0

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

8 y^{^{''}} - 9 y^{^{'}} + y = 0, y (0) = 8, y^{^{'}} (0) = \frac{1}{8}

y^{^{''''}} - 2 y^{^{'''}} - 10 y^{^{''}} + 6 y^{^{'}} + 45 y = 0

What is the solution for 4y^{''}-16y^'+17y=0 ?
The solution for 4y^{''}-16y^'+17y=0 is y=e^{2t}(c_{1}cos(t/2)+c_{2}sin(t/2))