What is the solution for y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=6 ?

The solution for y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=6 is y=e^{-t}(3cos(3t)+3sin(3t))

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y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=6

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (0) = 3, y^{^{'}} (0) = 6

y = e^{- t} (3 cos (3 t) + 3 sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (3 cos (3 t) + 3 sin (3 t))

y = e^{- t} (3 cos (3 t) + 3 sin (3 t))

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 8 x = 0, x = 0, t = 0

y^{^{''}} + a \cdot y = 0

y^{^{''}} - 4 c^{2} y = 0

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} + 4 y = 0

y^{^{'''}} - 2 y^{^{''}} - 11 y^{^{'}} + 12 y = 0

What is the solution for y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=6 ?
The solution for y^{''}+2y^'+10y=0,y(0)=3,y^'(0)=6 is y=e^{-t}(3cos(3t)+3sin(3t))