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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (x y + x cos (x))

x^{^{''}} + 9 x - 27 = 0, x (0) = 4, x^{^{'}} (0) = 6

\int \frac{1}{12 y} d y

\frac{\partial}{\partial y} (\frac{y - t}{5 y + 2 t})

t - π

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