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

The solution for y^{''}+2y^'+y=0,y(0)=-1 is y=-e^{-t}+c_{2}e^{-t}t

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

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

y = - e^{- t} + c_{2} e^{- t} t

Solution steps

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

Solve linear ODE:

y = - e^{- t} + c_{2} e^{- t} t

y = - e^{- t} + c_{2} e^{- t} t

\frac{d ^{2} y}{d x ^{2}} (1) - \frac{d y}{d x} (7) + 10 y = 0

y^{^{''}} + 4 y^{^{'}} + 20 y = 0, y (π) = 8, y (0) = 1

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = - 4, y^{^{'}} (0) = 14

9 \frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + y = 0

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

What is the solution for y^{''}+2y^'+y=0,y(0)=-1 ?
The solution for y^{''}+2y^'+y=0,y(0)=-1 is y=-e^{-t}+c_{2}e^{-t}t