What is the solution for 2y^{''}-6y^'+8y=0 ?

The solution for 2y^{''}-6y^'+8y=0 is y=e^{(3t)/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))

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2y^{''}-6y^'+8y=0

2 y^{^{''}} - 6 y^{^{'}} + 8 y = 0

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

2 y^{''} (t) - 6 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{\frac{3 t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

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

a \cdot y^{^{''}} + b \cdot y = 0

y^{^{''}} + 4 y^{^{'}} - 60 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = 4

y^{^{''}} + 8 y^{^{'}} + 15 y = 0, y (0) = 12, y^{^{'}} (0) = - 52

What is the solution for 2y^{''}-6y^'+8y=0 ?
The solution for 2y^{''}-6y^'+8y=0 is y=e^{(3t)/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))