What is the solution for 4y^{''}-2y^'+9y=0 ?

The solution for 4y^{''}-2y^'+9y=0 is y=e^{t/4}(c_{1}cos((sqrt(35)t)/4)+c_{2}sin((sqrt(35)t)/4))

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4y^{''}-2y^'+9y=0

4 y^{^{''}} - 2 y^{^{'}} + 9 y = 0

y = e^{\frac{t}{4}} (c_{1} cos (\frac{35 t}{4}) + c_{2} sin (\frac{35 t}{4}))

Solution steps

4 y^{''} (t) - 2 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{\frac{t}{4}} (c_{1} cos (\frac{35 t}{4}) + c_{2} sin (\frac{35 t}{4}))

y = e^{\frac{t}{4}} (c_{1} cos (\frac{35 t}{4}) + c_{2} sin (\frac{35 t}{4}))

y^{^{''}} + 19 y^{^{'}} + 90 y = 0

y^{^{'''}} - 4 y^{^{''}} - 41 y^{^{'}} - 36 y = 0

y^{^{'''}} + 10 y^{^{''}} - y^{^{'}} - 3 y = 0

y^{^{''}} + 9 y = 0, y (0) = 3, y^{^{'}} (0) = 12

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

What is the solution for 4y^{''}-2y^'+9y=0 ?
The solution for 4y^{''}-2y^'+9y=0 is y=e^{t/4}(c_{1}cos((sqrt(35)t)/4)+c_{2}sin((sqrt(35)t)/4))