What is the solution for y^{''}+10y^'+25y=0,y(-1)=2,y^'(-1)=3 ?

The solution for y^{''}+10y^'+25y=0,y(-1)=2,y^'(-1)=3 is y=15e^{-5t-5}+13e^{-5t-5}t

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y^{''}+10y^'+25y=0,y(-1)=2,y^'(-1)=3

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 3

y = 15 e^{- 5 t - 5} + 13 e^{- 5 t - 5} t

Solution steps

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

Solve linear ODE:

y = 15 e^{- 5 t - 5} + 13 e^{- 5 t - 5} t

y = 15 e^{- 5 t - 5} + 13 e^{- 5 t - 5} t

2 y^{^{''}} + 16 y^{^{'}} + 82 y = 0

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

121 y^{^{''}} + 198 y^{^{'}} + 80 y = 0

2 y^{^{'''}} + 7 y^{^{''}} + 4 y^{^{'}} - 4 y = 0

y^{^{''}} + 25 y^{^{'}} + 2500 y = 0

What is the solution for y^{''}+10y^'+25y=0,y(-1)=2,y^'(-1)=3 ?
The solution for y^{''}+10y^'+25y=0,y(-1)=2,y^'(-1)=3 is y=15e^{-5t-5}+13e^{-5t-5}t