What is the solution for x^{''}+2x^'+10x=0,x^'(0)=1,x(0)=0 ?

The solution for x^{''}+2x^'+10x=0,x^'(0)=1,x(0)=0 is x= 1/3 e^{-t}sin(3t)

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x^{''}+2x^'+10x=0,x^'(0)=1,x(0)=0

x^{^{''}} + 2 x^{^{'}} + 10 x = 0, x^{^{'}} (0) = 1, x (0) = 0

x = \frac{1}{3} e^{- t} sin (3 t)

Solution steps

x^{''} (t) + 2 x^{'} (t) + 10 x = 0

Solve linear ODE:

x = \frac{1}{3} e^{- t} sin (3 t)

x = \frac{1}{3} e^{- t} sin (3 t)

x^{^{''}} + 6 x^{^{'}} + 10 x = 0

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = 3, y^{^{'}} (0) = 7

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 16 y = 0, y (0) = 5

y^{^{''''}} + 4 y^{^{'''}} + 13 y^{^{''}} + 36 y^{^{'}} + 36 y = 0

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

What is the solution for x^{''}+2x^'+10x=0,x^'(0)=1,x(0)=0 ?
The solution for x^{''}+2x^'+10x=0,x^'(0)=1,x(0)=0 is x= 1/3 e^{-t}sin(3t)