What is the solution for y^{''}+3y^'+2.25y=0 ?

The solution for y^{''}+3y^'+2.25y=0 is y=c_{1}e^{-(3t)/2}+c_{2}e^{-(3t)/2}t

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y^{''}+3y^'+2.25y=0

y^{^{''}} + 3 y^{^{'}} + 2.25 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) + 2.25 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

\frac{d}{d x} (5 e^{x} + \frac{4}{3 x})

\frac{d y}{d t} = 7 y^{2} t^{3}, y (2) = 3

\int_{1}^{3} \frac{t}{3 + t ^{2}} d t

\int \frac{2 x + 3}{x + 2} d x

\frac{d}{d t} (a t - tanh (a t))

What is the solution for y^{''}+3y^'+2.25y=0 ?
The solution for y^{''}+3y^'+2.25y=0 is y=c_{1}e^{-(3t)/2}+c_{2}e^{-(3t)/2}t