What is the solution for 2y^{''}+3y^'+4y=0 ?

The solution for 2y^{''}+3y^'+4y=0 is y=e^{-(3t)/4}(c_{1}cos((sqrt(23)t)/4)+c_{2}sin((sqrt(23)t)/4))

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2y^{''}+3y^'+4y=0

2 y^{^{''}} + 3 y^{^{'}} + 4 y = 0

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

Solution steps

2 y^{''} (t) + 3 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

y = e^{- \frac{3 t}{4}} (c_{1} cos (\frac{23 t}{4}) + c_{2} sin (\frac{23 t}{4}))

What is the solution for 2y^{''}+3y^'+4y=0 ?
The solution for 2y^{''}+3y^'+4y=0 is y=e^{-(3t)/4}(c_{1}cos((sqrt(23)t)/4)+c_{2}sin((sqrt(23)t)/4))