What is the solution for 4y^{''}+28y^'+49y=0 ?

The solution for 4y^{''}+28y^'+49y=0 is y=c_{1}e^{-(7t)/2}+c_{2}e^{-(7t)/2}t

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4y^{''}+28y^'+49y=0

4 y^{^{''}} + 28 y^{^{'}} + 49 y = 0

y = c_{1} e^{- \frac{7 t}{2}} + c_{2} e^{- \frac{7 t}{2}} t

Solution steps

4 y^{''} (t) + 28 y^{'} (t) + 49 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{7 t}{2}} + c_{2} e^{- \frac{7 t}{2}} t

y = c_{1} e^{- \frac{7 t}{2}} + c_{2} e^{- \frac{7 t}{2}} t

f (t) = e^{t + 9}

x \to + 1 lim (\frac{4}{x - 1})

\frac{d}{d x} (e^{- 4 x} - 4 e^{- 4 x} x)

\frac{d}{d x} (ln (\frac{( x + y )}{( x - y )}))

\frac{d}{d x} (e^{x^{9}})

What is the solution for 4y^{''}+28y^'+49y=0 ?
The solution for 4y^{''}+28y^'+49y=0 is y=c_{1}e^{-(7t)/2}+c_{2}e^{-(7t)/2}t