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

The solution for 4y^{''}+2y^'+2y=0 is y=e^{-t/4}(c_{1}cos((sqrt(7)t)/4)+c_{2}sin((sqrt(7)t)/4))

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4y^{''}+2y^'+2y=0

4 y^{^{''}} + 2 y^{^{'}} + 2 y = 0

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

y = cos (x), at x = \frac{π}{2}

\frac{d}{d x} (x + x x)

\frac{d}{d x} (x^{3} + 3^{x})

\int sin (x) 2 x d x

x \to \infty lim (\frac{( x ^{2} )}{e ^{x^{2}}})

What is the solution for 4y^{''}+2y^'+2y=0 ?
The solution for 4y^{''}+2y^'+2y=0 is y=e^{-t/4}(c_{1}cos((sqrt(7)t)/4)+c_{2}sin((sqrt(7)t)/4))