What is the solution for u^2y^{''}-4uy^'+6y=0 ?

The solution for u^2y^{''}-4uy^'+6y=0 is y=e^{(2t)/u}(c_{1}cos((sqrt(2)t)/u)+c_{2}sin((sqrt(2)t)/u))

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u^2y^{''}-4uy^'+6y=0

u^{2} y^{^{''}} - 4 u y^{^{'}} + 6 y = 0

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

Solution steps

u^{2} y^{''} (t) - 4 u y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

9 y^{^{''}} + 30 y = 0, y (0) = 8, y^{^{'}} (0) = 8

50 x^{^{''}} + 100 x^{^{'}} + 1000 x = 0

\frac{1}{10} y^{^{''}} + 11 y^{^{'}} + 24 y = 0, y (0) = 1, y^{^{'}} (0) = 0

q^{^{''}} + 6 q^{^{'}} + 9 q = 0

0.1 y^{^{''}} + 4 y^{^{'}} + 250 y = 0

What is the solution for u^2y^{''}-4uy^'+6y=0 ?
The solution for u^2y^{''}-4uy^'+6y=0 is y=e^{(2t)/u}(c_{1}cos((sqrt(2)t)/u)+c_{2}sin((sqrt(2)t)/u))