What is the solution for y^{''}-3y^'+3y=0 ?

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

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

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

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

Solution steps

y^{''} (t) - 3 y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

x^{^{''}} - 2 x^{^{'}} + 2 x = 0, x (0) = 1, x^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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

y^{^{''}} + 49 y = 0, y (π) = 0, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} + a y = 0

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