What is the solution for y^{''}+y^'+4.25y=0,y(0)=5 ?

The solution for y^{''}+y^'+4.25y=0,y(0)=5 is y=e^{-t/2}(5cos(2t)+c_{2}sin(2t))

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y^{''}+y^'+4.25y=0,y(0)=5

y^{^{''}} + y^{^{'}} + 4.25 y = 0, y (0) = 5

y = e^{- \frac{t}{2}} (5 cos (2 t) + c_{2} sin (2 t))

Solution steps

y^{''} (t) + y^{'} (t) + 4.25 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (5 cos (2 t) + c_{2} sin (2 t))

y = e^{- \frac{t}{2}} (5 cos (2 t) + c_{2} sin (2 t))

What is the solution for y^{''}+y^'+4.25y=0,y(0)=5 ?
The solution for y^{''}+y^'+4.25y=0,y(0)=5 is y=e^{-t/2}(5cos(2t)+c_{2}sin(2t))