What is the solution for 2y^{''}+4y^'+10y=0,y(0)=1,y^'(0)=-2 ?

The solution for 2y^{''}+4y^'+10y=0,y(0)=1,y^'(0)=-2 is y=e^{-t}(cos(2t)-1/2 sin(2t))

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2y^{''}+4y^'+10y=0,y(0)=1,y^'(0)=-2

2 y^{^{''}} + 4 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

y = e^{- t} (cos (2 t) - \frac{1}{2} sin (2 t))

Solution steps

2 y^{''} (t) + 4 y^{'} (t) + 10 y = 0

Solve linear ODE:

y = e^{- t} (cos (2 t) - \frac{1}{2} sin (2 t))

y = e^{- t} (cos (2 t) - \frac{1}{2} sin (2 t))

5 y^{^{'''}} + 11 y^{^{''}} - 38 y^{^{'}} - 8 y = 0

4 \frac{d ^{2} x}{d t ^{2}} + 324 x = 0

y^{^{'''}} - 20 y^{^{''}} + 111 y^{^{'}} - 180 y = 0

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 0

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (1) = 5

What is the solution for 2y^{''}+4y^'+10y=0,y(0)=1,y^'(0)=-2 ?
The solution for 2y^{''}+4y^'+10y=0,y(0)=1,y^'(0)=-2 is y=e^{-t}(cos(2t)-1/2 sin(2t))