What is the solution for y^{''}+8y^'+25y=0,y(0)=2,y^'(0)=-7 ?

The solution for y^{''}+8y^'+25y=0,y(0)=2,y^'(0)=-7 is y=e^{-4t}(2cos(3t)+1/3 sin(3t))

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y^{''}+8y^'+25y=0,y(0)=2,y^'(0)=-7

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 7

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 25 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 0.1 y = 0

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

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

y^{^{'''}} + 9 y^{^{'}} = 0, y^{^{'}} (0) = - 21

16 y^{^{''}} + 8 y^{^{'}} + 145 y = 0

What is the solution for y^{''}+8y^'+25y=0,y(0)=2,y^'(0)=-7 ?
The solution for y^{''}+8y^'+25y=0,y(0)=2,y^'(0)=-7 is y=e^{-4t}(2cos(3t)+1/3 sin(3t))