What is the solution for 4y^{''}+7y^'+8y=0 ?

The solution for 4y^{''}+7y^'+8y=0 is y=e^{-(7t)/8}(c_{1}cos((sqrt(79)t)/8)+c_{2}sin((sqrt(79)t)/8))

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4y^{''}+7y^'+8y=0

4 y^{^{''}} + 7 y^{^{'}} + 8 y = 0

y = e^{- \frac{7 t}{8}} (c_{1} cos (\frac{79 t}{8}) + c_{2} sin (\frac{79 t}{8}))

Solution steps

4 y^{''} (t) + 7 y^{'} (t) + 8 y = 0

Solve linear ODE:

y = e^{- \frac{7 t}{8}} (c_{1} cos (\frac{79 t}{8}) + c_{2} sin (\frac{79 t}{8}))

y = e^{- \frac{7 t}{8}} (c_{1} cos (\frac{79 t}{8}) + c_{2} sin (\frac{79 t}{8}))

What is the solution for 4y^{''}+7y^'+8y=0 ?
The solution for 4y^{''}+7y^'+8y=0 is y=e^{-(7t)/8}(c_{1}cos((sqrt(79)t)/8)+c_{2}sin((sqrt(79)t)/8))