What is the solution for y^{''}+2y^'+y=0,y^'(0)=5 ?

The solution for y^{''}+2y^'+y=0,y^'(0)=5 is y=(-5+c_{2})e^{-t}+c_{2}te^{-t}

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

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

y = (- 5 + c_{2}) e^{- t} + c_{2} t e^{- t}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 0

Solve linear ODE:

y = (- 5 + c_{2}) e^{- t} + c_{2} t e^{- t}

y = (- 5 + c_{2}) e^{- t} + c_{2} t e^{- t}

16 y^{^{''}} - 8 y^{^{'}} + y = 0, y (0) = 40, y^{^{'}} (0) = - 10

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

(D^{4} - 6 D^{3} + 5 D^{2}) y = 0

y^{^{'''}} - 24 y^{^{''}} + 192 y^{^{'}} - 512 y = 0

y^{^{''}} + 4 \cdot π \cdot \frac{y ^{^{'}}}{20} + 100 \cdot π^{2} \cdot y = 0

What is the solution for y^{''}+2y^'+y=0,y^'(0)=5 ?
The solution for y^{''}+2y^'+y=0,y^'(0)=5 is y=(-5+c_{2})e^{-t}+c_{2}te^{-t}