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

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

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

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

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

Solution steps

y^{'} (t) + 2 y^{''} (t) + 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^{^{''}} - (8) y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 73

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 0, y^{^{'}} (0) = 1

9 y^{^{''}} + 18 y^{^{'}} + 13 y = 0

q^{^{''}} + 20 q^{^{'}} + 100 q = 0

\frac{d ^{2} y}{d x ^{2}} - 366 \frac{d y}{d x} = 0

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