What is the solution for 4y^{''}+4y^'+325y=0,y(0)=3,y^'(0)=2 ?

The solution for 4y^{''}+4y^'+325y=0,y(0)=3,y^'(0)=2 is y=e^{-t/2}(3cos(9t)+7/18 sin(9t))

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4y^{''}+4y^'+325y=0,y(0)=3,y^'(0)=2

4 y^{^{''}} + 4 y^{^{'}} + 325 y = 0, y (0) = 3, y^{^{'}} (0) = 2

y = e^{- \frac{t}{2}} (3 cos (9 t) + \frac{7}{18} sin (9 t))

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + 325 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (3 cos (9 t) + \frac{7}{18} sin (9 t))

y = e^{- \frac{t}{2}} (3 cos (9 t) + \frac{7}{18} sin (9 t))

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

\frac{d ^{2} y}{d x ^{2}} - 25 \cdot y = 0

x^{^{''}} = - x^{^{'}} + 3 x

y^{^{'''}} - 7 y^{^{''}} + 14 y^{^{'}} - 8 y = 0

y^{^{''}} + 2 y^{^{'}} + y = - y^{^{''}} + 5 y + 2 y^{^{'}} - y

What is the solution for 4y^{''}+4y^'+325y=0,y(0)=3,y^'(0)=2 ?
The solution for 4y^{''}+4y^'+325y=0,y(0)=3,y^'(0)=2 is y=e^{-t/2}(3cos(9t)+7/18 sin(9t))