What is the solution for y^{''}=(-y^'-5y)/4 ?

The solution for y^{''}=(-y^'-5y)/4 is y=e^{-t/8}(c_{1}cos((sqrt(79)t)/8)+c_{2}sin((sqrt(79)t)/8))

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y^{''}=(-y^'-5y)/4

y^{^{''}} = \frac{- y ^{^{'}} - 5 y}{4}

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{79 t}{8}) + c_{2} sin (\frac{79 t}{8}))

Solution steps

y^{''} (t) = \frac{- y ^{'} ( t ) - 5 y}{4}

Solve linear ODE:

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{79 t}{8}) + c_{2} sin (\frac{79 t}{8}))

y = e^{- \frac{t}{8}} (c_{1} cos (\frac{79 t}{8}) + c_{2} sin (\frac{79 t}{8}))

f^{^{''}} (x) = - f (x)

y^{^{''}} - n^{2} π^{2} y = 0

y^{^{''}} + 2 y^{^{''}} - 3 y = 0

y^{^{''}} - y = 0, y (0) = \frac{1}{5}, y (1) = \frac{1}{5} cosh (1)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 0

What is the solution for y^{''}=(-y^'-5y)/4 ?
The solution for y^{''}=(-y^'-5y)/4 is y=e^{-t/8}(c_{1}cos((sqrt(79)t)/8)+c_{2}sin((sqrt(79)t)/8))