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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

\frac{d}{d x} (x^{3} + 4 x^{2} - 6 x + 1)

20 y^{^{''}} - 11 y^{^{'}} - 3 y = 0

\int 5 sin^{3} (5 x co) s^{3} 5 x d x

\int_{9}^{10} \frac{1}{( x - 9 ) ^{\frac{5}{4}}} d x

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