What is the solution for y^{''}+8y^'+25y=0,y(0)=1,y^'(0)=2 ?

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

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y^{''}+8y^'+25y=0,y(0)=1,y^'(0)=2

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 2

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

Solution steps

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

Solve linear ODE:

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

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

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

2 y^{^{''}} + 8 y + 10 y = 0, y (0) = 5, y^{^{'}} (0) = 0

5 x^{^{''}} + 10 x^{^{'}} + 25 x = 0

u^{^{''}} + 7 u^{^{'}} + 2 H (t) = 0

y^{^{''}} + 6 y^{^{'}} - 16 y = 7 y^{^{'}} - 10 y

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