What is the solution for y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-3 ?

The solution for y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-3 is y=4e^{-t}+e^{-t}t

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y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-3

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 4, y^{^{'}} (0) = - 3

y = 4 e^{- t} + e^{- t} t

Solution steps

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

Solve linear ODE:

y = 4 e^{- t} + e^{- t} t

y = 4 e^{- t} + e^{- t} t

\frac{d}{d x} (\frac{x ^{2} - 3 x + 1}{x ^{2}})

x \to - 1 lim (8 x^{4} - 3 x^{3} + 7 x)

\frac{d}{d x} (sin (x^{2} + 2 x - 1))

\frac{\partial}{\partial x} (ln (x + y))

f (u) = u^{2}

What is the solution for y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-3 ?
The solution for y^{''}+2y^'+y=0,y(0)=4,y^'(0)=-3 is y=4e^{-t}+e^{-t}t