What is the solution for y^{''}+3y^'+7y=0 ?

The solution for y^{''}+3y^'+7y=0 is y=e^{-(3t)/2}(c_{1}cos((sqrt(19)t)/2)+c_{2}sin((sqrt(19)t)/2))

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y^{''}+3y^'+7y=0

y^{^{''}} + 3 y^{^{'}} + 7 y = 0

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

Solution steps

y^{''} (t) + 3 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

y = e^{- \frac{3 t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

y^{^{''}} + y^{^{'}} = y

y^{^{'''}} + 2 y^{^{''}} - 19 y^{^{'}} - 20 y = 0

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 14 y = 0

y^{^{''}} - 13 y^{^{'}} + 30 y = 0

y^{^{'''}} - 5 y^{^{''}} - y^{^{'}} + 5 y = 0

What is the solution for y^{''}+3y^'+7y=0 ?
The solution for y^{''}+3y^'+7y=0 is y=e^{-(3t)/2}(c_{1}cos((sqrt(19)t)/2)+c_{2}sin((sqrt(19)t)/2))