What is the solution for 4y^{''}+12y^'+9y=0 ?

The solution for 4y^{''}+12y^'+9y=0 is y=c_{1}e^{-(3t)/2}+c_{2}e^{-(3t)/2}t

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4y^{''}+12y^'+9y=0

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0

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

Solution steps

4 y^{''} (t) + 12 y^{'} (t) + 9 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^{^{'}} = \frac{y}{x} + 2 x + 1

f (x) = \frac{1}{2} x^{8}

\frac{\partial}{\partial x} (y e^{x} + x e^{y})

t

(1.5) (2.8)

What is the solution for 4y^{''}+12y^'+9y=0 ?
The solution for 4y^{''}+12y^'+9y=0 is y=c_{1}e^{-(3t)/2}+c_{2}e^{-(3t)/2}t