What is the solution for 4y^{''}-7y^'+4y=0 ?

The solution for 4y^{''}-7y^'+4y=0 is y=e^{(7t)/8}(c_{1}cos((sqrt(15)t)/8)+c_{2}sin((sqrt(15)t)/8))

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4y^{''}-7y^'+4y=0

4 y^{^{''}} - 7 y^{^{'}} + 4 y = 0

y = e^{\frac{7 t}{8}} (c_{1} cos (\frac{15 t}{8}) + c_{2} sin (\frac{15 t}{8}))

Solution steps

4 y^{''} (t) - 7 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{\frac{7 t}{8}} (c_{1} cos (\frac{15 t}{8}) + c_{2} sin (\frac{15 t}{8}))

y = e^{\frac{7 t}{8}} (c_{1} cos (\frac{15 t}{8}) + c_{2} sin (\frac{15 t}{8}))

y^{^{''}} + y = 0, y (\frac{π}{3}) = 0, y^{^{'}} (\frac{π}{3}) = 2

y^{^{''}} + 3 y^{^{'}} + 6 y = 0

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

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

(D^{3} + D^{2} + 4 D + 4) y = 0

What is the solution for 4y^{''}-7y^'+4y=0 ?
The solution for 4y^{''}-7y^'+4y=0 is y=e^{(7t)/8}(c_{1}cos((sqrt(15)t)/8)+c_{2}sin((sqrt(15)t)/8))