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

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

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y^{''}+2y^'+1y=0

y^{^{''}} + 2 y^{^{'}} + 1 y = 0

y = c_{1} e^{- t} + c_{2} t e^{- t}

Solution steps

y^{''} (t) + 2 y^{'} (t) + 1 y = 0

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t}

\frac{\partial}{\partial y} (\frac{x}{( 1 + 2 x + 3 y )})

2 y^{^{''}} - 9 y^{^{'}} + 4 y = 0

f (x) = x^{6} e^{x}

y = e^{- 2 x} \cdot tan (8 x)

x \to 0 lim (\frac{3 sin ( 3 x )}{x})

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