What is the solution for y^{''}+4y^'+7y=0 ?

The solution for y^{''}+4y^'+7y=0 is y=e^{-2t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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y^{''}+4y^'+7y=0

y^{^{''}} + 4 y^{^{'}} + 7 y = 0

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{- 2 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

\frac{\partial}{\partial x} (x ln (y + x))

y^{^{'}} = e (- 2 t) - y

f (x) = (x^{3} + 5 x^{2} - 3) x

(cos (3 x))^{^{'}}

\frac{d}{d x} (e^{x} - 1 - x)

What is the solution for y^{''}+4y^'+7y=0 ?
The solution for y^{''}+4y^'+7y=0 is y=e^{-2t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))