What is the solution for y^{''}+6y^'+9y=0,y(1)=0 ?

The solution for y^{''}+6y^'+9y=0,y(1)=0 is y=-c_{2}e^{-3t}+c_{2}e^{-3t}t

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y^{''}+6y^'+9y=0,y(1)=0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (1) = 0

y = - c_{2} e^{- 3 t} + c_{2} e^{- 3 t} t

Solution steps

y^{''} (t) + 6 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = - c_{2} e^{- 3 t} + c_{2} e^{- 3 t} t

y = - c_{2} e^{- 3 t} + c_{2} e^{- 3 t} t

\int sin^{2} (x) \cdot cos (x) d x

f (x) = (x^{2} - 1) (- 3 x^{3} + 2 x)

\int (cos^{2} (x) sin (x)) d x

\frac{\partial}{\partial x} (- \frac{1}{a x ^{2}})

x = 18.432 t cos (1.44 t^{2})

What is the solution for y^{''}+6y^'+9y=0,y(1)=0 ?
The solution for y^{''}+6y^'+9y=0,y(1)=0 is y=-c_{2}e^{-3t}+c_{2}e^{-3t}t