What is the solution for x^{''}+2x^'+9x=0,x(0)=0,x^'(0)=1 ?

The solution for x^{''}+2x^'+9x=0,x(0)=0,x^'(0)=1 is x=(e^{-t}sin(2^{1/2+1}t))/(2sqrt(2))

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x^{''}+2x^'+9x=0,x(0)=0,x^'(0)=1

x^{^{''}} + 2 x^{^{'}} + 9 x = 0, x (0) = 0, x^{^{'}} (0) = 1

x = \frac{e ^{- t} sin ( 2 ^{\frac{1}{2} + 1} t )}{2 2}

Solution steps

x^{''} (t) + 2 x^{'} (t) + 9 x = 0

Solve linear ODE:

x = \frac{e ^{- t} sin ( 2 ^{\frac{1}{2} + 1} t )}{2 2}

x = \frac{e ^{- t} sin ( 2 ^{\frac{1}{2} + 1} t )}{2 2}

\frac{d ^{2} y}{d x ^{2}} - 1 \frac{d y}{d x} = 0

y^{^{''}} + 12 y^{^{'}} + 16 y = 0

y^{^{''}} - y - 12 y = 0

f^{^{''}} (x) + f^{^{'}} (x) = 0

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

What is the solution for x^{''}+2x^'+9x=0,x(0)=0,x^'(0)=1 ?
The solution for x^{''}+2x^'+9x=0,x(0)=0,x^'(0)=1 is x=(e^{-t}sin(2^{1/2+1}t))/(2sqrt(2))