What is the solution for y^{''}+6y^'+10y=0,y(0)=2,y^'(0)=1 ?

The solution for y^{''}+6y^'+10y=0,y(0)=2,y^'(0)=1 is y=e^{-3t}(2cos(t)+7sin(t))

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y^{''}+6y^'+10y=0,y(0)=2,y^'(0)=1

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = e^{- 3 t} (2 cos (t) + 7 sin (t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{- 3 t} (2 cos (t) + 7 sin (t))

y = e^{- 3 t} (2 cos (t) + 7 sin (t))

y^{^{''}} - 8 y^{^{'}} + 17 y = 0, y (0) = 2, y^{^{'}} (0) = 10

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

y^{^{''}} + 4 y^{^{'}} + 16 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d t ^{2}} + 2 y = 0, y^{^{'}} (0) = 2

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

What is the solution for y^{''}+6y^'+10y=0,y(0)=2,y^'(0)=1 ?
The solution for y^{''}+6y^'+10y=0,y(0)=2,y^'(0)=1 is y=e^{-3t}(2cos(t)+7sin(t))