What is the solution for y^{''}+6y^'+9y=0,y(0)=e^{-3x} ?

The solution for y^{''}+6y^'+9y=0,y(0)=e^{-3x} is y=c_{1}e^{-3x}+c_{2}xe^{-3x}

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y^{''}+6y^'+9y=0,y(0)=e^{-3x}

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = e^{- 3 x}

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 9 y = 0

Solve linear ODE:

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x}

y = c_{1} e^{- 3 x} + c_{2} x e^{- 3 x}

2 y^{^{''}} - 7 y^{^{'}} + 5 y = 0, y (0) = 4, y^{^{'}} (0) = 5.5

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

\frac{d ^{2} y}{d t ^{2}} + 18 \frac{d y}{d t} + 81 y = 0

6 y^{^{''}} + 120 y^{^{'}} + 50 y = 0

\frac{d ^{2} q}{d t ^{2}} + q^{^{'}} + q = 0

What is the solution for y^{''}+6y^'+9y=0,y(0)=e^{-3x} ?
The solution for y^{''}+6y^'+9y=0,y(0)=e^{-3x} is y=c_{1}e^{-3x}+c_{2}xe^{-3x}