What is the solution for y^{''}+10y^'+25y=0,y(0)=-5,y^'(0)=28 ?

The solution for y^{''}+10y^'+25y=0,y(0)=-5,y^'(0)=28 is y=-5e^{-5t}+3e^{-5t}t

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y^{''}+10y^'+25y=0,y(0)=-5,y^'(0)=28

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 5, y^{^{'}} (0) = 28

y = - 5 e^{- 5 t} + 3 e^{- 5 t} t

Solution steps

y^{''} (t) + 10 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = - 5 e^{- 5 t} + 3 e^{- 5 t} t

y = - 5 e^{- 5 t} + 3 e^{- 5 t} t

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 13

16 y^{^{'''}} + 24 y^{^{''}} + 9 y^{^{'}} = 0

121 y^{^{''}} - 4 y = 0

64 y^{^{''}} + 176 y^{^{'}} + 121 y = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0

What is the solution for y^{''}+10y^'+25y=0,y(0)=-5,y^'(0)=28 ?
The solution for y^{''}+10y^'+25y=0,y(0)=-5,y^'(0)=28 is y=-5e^{-5t}+3e^{-5t}t