What is the solution for y^{''}+10y^'+25y=0,y(0)=-4,y^'(0)=18 ?

The solution for y^{''}+10y^'+25y=0,y(0)=-4,y^'(0)=18 is y=-4e^{-5t}-2e^{-5t}t

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y^{''}+10y^'+25y=0,y(0)=-4,y^'(0)=18

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (0) = - 4, y^{^{'}} (0) = 18

y = - 4 e^{- 5 t} - 2 e^{- 5 t} t

Solution steps

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

Solve linear ODE:

y = - 4 e^{- 5 t} - 2 e^{- 5 t} t

y = - 4 e^{- 5 t} - 2 e^{- 5 t} t

6 y^{^{''}} + 108 y^{^{'}} + 486 y = 0

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y (0) = \frac{3}{2}, y^{^{'}} (0) = \frac{5}{2}

y^{^{'''}} + 19 y^{^{''}} + 93 y^{^{'}} + 75 y = 0

2 y^{^{'''}} - 5 y^{^{''}} - y^{^{'}} + 6 y = 0

y^{^{''}} + 10 y^{^{'}} + 15 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}+10y^'+25y=0,y(0)=-4,y^'(0)=18 ?
The solution for y^{''}+10y^'+25y=0,y(0)=-4,y^'(0)=18 is y=-4e^{-5t}-2e^{-5t}t