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

The solution for 4y^{''}-2y^'+y=0 is y=e^{t/4}(c_{1}cos((sqrt(3)t)/4)+c_{2}sin((sqrt(3)t)/4))

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

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

y = e^{\frac{t}{4}} (c_{1} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4}))

Solution steps

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

Solve linear ODE:

y = e^{\frac{t}{4}} (c_{1} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4}))

y = e^{\frac{t}{4}} (c_{1} cos (\frac{3 t}{4}) + c_{2} sin (\frac{3 t}{4}))

16 y^{^{''}} - 8 y^{^{'}} + 145 y = 0, y (0) = 1

16 \cdot y^{^{''}} + 24 \cdot y^{^{'}} + 73 \cdot y = 0, y (0) = [4]

64 y^{^{''}} - 81 y = 0

y^{^{''}} + 4 y = 0, y (3 π) = - 3, y (0) = - 3

y^{^{''}} - y^{^{'}} = 0, y (0) = 0, y^{^{'}} (0) = 0

What is the solution for 4y^{''}-2y^'+y=0 ?
The solution for 4y^{''}-2y^'+y=0 is y=e^{t/4}(c_{1}cos((sqrt(3)t)/4)+c_{2}sin((sqrt(3)t)/4))