What is the solution for y^{''}+3y^'+9/4 y=0 ?

The solution for y^{''}+3y^'+9/4 y=0 is y=c_{1}e^{-(3t)/2}+c_{2}e^{-(3t)/2}t

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y^{''}+3y^'+9/4 y=0

y^{^{''}} + 3 y^{^{'}} + \frac{9}{4} y = 0

y = c_{1} e^{- \frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}} t

Solution steps

y^{''} (t) + 3 y^{'} (t) + \frac{9}{4} y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}} t

y = c_{1} e^{- \frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}} t

y^{^{''}} - y^{^{'}} - 12 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

24 y^{^{''}} + 7 y^{^{'}} - 5 y = 0

2 y^{^{''}} + 12 y^{^{'}} + 18 y = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y^{^{'}} (0) = - 2

x^{^{''}} + 42 x^{^{'}} + 8 x = 0

What is the solution for y^{''}+3y^'+9/4 y=0 ?
The solution for y^{''}+3y^'+9/4 y=0 is y=c_{1}e^{-(3t)/2}+c_{2}e^{-(3t)/2}t