What is the solution for y^{''}+4y^'+(3^2+4^2)y=0 ?

The solution for y^{''}+4y^'+(3^2+4^2)y=0 is y=e^{-2t}(c_{1}cos(sqrt(21)t)+c_{2}sin(sqrt(21)t))

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y^{''}+4y^'+(3^2+4^2)y=0

y^{^{''}} + 4 y^{^{'}} + (3^{2} + 4^{2}) y = 0

y = e^{- 2 t} (c_{1} cos (21 t) + c_{2} sin (21 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + (3^{2} + 4^{2}) y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (21 t) + c_{2} sin (21 t))

y = e^{- 2 t} (c_{1} cos (21 t) + c_{2} sin (21 t))

y^{^{''}} + 2 y + y = 0, y^{1} (0) = x e^{- x}

3 y^{^{''}} - 12 y^{^{'}} - 36 y = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (\frac{1}{6}) = 2, y (0) = 3

16 x^{^{''}} + 40 x^{^{'}} + 169 x = 0

y^{^{''}} + 5 y^{^{'}} + 16 y = 0

What is the solution for y^{''}+4y^'+(3^2+4^2)y=0 ?
The solution for y^{''}+4y^'+(3^2+4^2)y=0 is y=e^{-2t}(c_{1}cos(sqrt(21)t)+c_{2}sin(sqrt(21)t))