What is the solution for 49y^{''}+14y^'+y=0 ?

The solution for 49y^{''}+14y^'+y=0 is y=c_{1}e^{-t/7}+c_{2}e^{-t/7}t

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49y^{''}+14y^'+y=0

49 y^{^{''}} + 14 y^{^{'}} + y = 0

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

Solution steps

49 y^{''} (t) + 14 y^{'} (t) + y = 0

Solve linear ODE:

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

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

\frac{d}{d x} ((2 x^{2} + 1)^{\frac{3}{2}})

\int_{- 2}^{1} (8 x^{3} - 3 x^{2}) d x

\frac{d y}{d x} = 3 x e^{y}

\frac{d}{d x} (- \frac{1}{2} cos (x))

(7 - 6 x)^{\frac{1}{2}}, at x = - 1

What is the solution for 49y^{''}+14y^'+y=0 ?
The solution for 49y^{''}+14y^'+y=0 is y=c_{1}e^{-t/7}+c_{2}e^{-t/7}t