What is the solution for 36y^{''}+108y^'+97y=0,y(1)=4 ?

The solution for 36y^{''}+108y^'+97y=0,y(1)=4 is y=e^{-(3t)/2}(c_{1}cos((2t)/3)+c_{2}sin((2t)/3))

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36y^{''}+108y^'+97y=0,y(1)=4

36 y^{''} + 108 y^{'} + 97 y = 0

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

Solution steps

36 y^{''} (t) + 108 y^{'} (t) + 97 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y (1) = 6

y^{^{'''}} + 9 y^{^{''}} + 15 y^{^{'}} - 25 y = 0

y^{^{'''}} - y^{^{''}} - 18 y^{^{'}} - 10 y = 0

y^{^{''''}} + 7 y^{^{'''}} + 14 y^{^{''}} - 22 y^{^{'}} - 84 y = 0

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = - 8

What is the solution for 36y^{''}+108y^'+97y=0,y(1)=4 ?
The solution for 36y^{''}+108y^'+97y=0,y(1)=4 is y=e^{-(3t)/2}(c_{1}cos((2t)/3)+c_{2}sin((2t)/3))