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

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

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y^{''}+2y^'+10y=0,y(0)=-2,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = - 2, y^{^{'}} (0) = 0

y = e^{- t} (- 2 cos (3 t) - \frac{2}{3} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (- 2 cos (3 t) - \frac{2}{3} sin (3 t))

y = e^{- t} (- 2 cos (3 t) - \frac{2}{3} sin (3 t))

\frac{d ^{2}}{d θ ^{2}} (y (θ)) + y (θ) = 0

x^{^{''}} + 2 x^{^{'}} + 10 x = 0

y^{^{'''}} - 3 y^{^{''}} - 4 y^{^{'}} + 12 y = 0

y^{^{''}} + y = 0, y (0) = sin (x)

9 \frac{d ^{2} y}{d t ^{2}} - 24 \frac{d y}{d t} + 16 y = 0

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