What is the solution for y^{''}+6y^'+25y=0,y(0)= 1/2 ,y^'(0)=0 ?

The solution for y^{''}+6y^'+25y=0,y(0)= 1/2 ,y^'(0)=0 is y=e^{-3t}(1/2 cos(4t)+3/8 sin(4t))

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y^{''}+6y^'+25y=0,y(0)= 1/2 ,y^'(0)=0

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 0

y = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

y = e^{- 3 t} (\frac{1}{2} cos (4 t) + \frac{3}{8} sin (4 t))

\frac{d ^{2} y}{d x ^{2}} - 9 \frac{d y}{d x} + 18 y = 0

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y^{^{'}} (0) = 1, y (0) = 1

y^{^{''}} + 1000 y^{^{'}} + 250000 y = 0

y^{^{''}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = - 7

2 y - y^{^{''}} = 0

What is the solution for y^{''}+6y^'+25y=0,y(0)= 1/2 ,y^'(0)=0 ?
The solution for y^{''}+6y^'+25y=0,y(0)= 1/2 ,y^'(0)=0 is y=e^{-3t}(1/2 cos(4t)+3/8 sin(4t))