What is the solution for 4y^{''}-12y^'+10y=0 ?

The solution for 4y^{''}-12y^'+10y=0 is y=e^{(3t)/2}(c_{1}cos(t/2)+c_{2}sin(t/2))

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4y^{''}-12y^'+10y=0

4 y^{^{''}} - 12 y^{^{'}} + 10 y = 0

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

Solution steps

4 y^{''} (t) - 12 y^{'} (t) + 10 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} + 13 y = 0

y^{^{''}} + 10 y^{^{'}} + 23 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 15 y^{^{'}} + 50 y = 0, y (0) = 8, y^{^{'}} (0) = - 55

y^{^{''}} + 5 y^{^{'}} + 25 y = 0

y^{^{''}} - 14 y^{^{'}} + 49 y = 0, y (0) = - 6, y^{^{'}} (0) = 0

What is the solution for 4y^{''}-12y^'+10y=0 ?
The solution for 4y^{''}-12y^'+10y=0 is y=e^{(3t)/2}(c_{1}cos(t/2)+c_{2}sin(t/2))