What is the solution for x^{''}+1/4 x^'+x=0 ?

The solution for x^{''}+1/4 x^'+x=0 is x=e^{-t/8}(c_{1}cos((3sqrt(7)t)/8)+c_{2}sin((3sqrt(7)t)/8))

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x^{''}+1/4 x^'+x=0

x^{^{''}} + \frac{1}{4} x^{^{'}} + x = 0

x = e^{- \frac{t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

Solution steps

x^{''} (t) + \frac{1}{4} x^{'} (t) + x = 0

Solve linear ODE:

x = e^{- \frac{t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

x = e^{- \frac{t}{8}} (c_{1} cos (\frac{3 7 t}{8}) + c_{2} sin (\frac{3 7 t}{8}))

\frac{d ^{2} x}{d t ^{2}} = - k \frac{d x}{d t}

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 2

x^{^{''}} + \frac{32}{2} x = 0, x (0) = - 1, x^{^{'}} (0) = - 5

\frac{d ^{2} y}{d x ^{2}} + 9 \frac{d y}{d x} + 8 y = 0

2 y^{^{''}} + 12 y^{^{'}} + 68 y = 0, y (0) = 3, y^{^{'}} (0) = 0

What is the solution for x^{''}+1/4 x^'+x=0 ?
The solution for x^{''}+1/4 x^'+x=0 is x=e^{-t/8}(c_{1}cos((3sqrt(7)t)/8)+c_{2}sin((3sqrt(7)t)/8))