What is the solution for y^{''}+6y^'+10y=0,y(0)=4 ?

The solution for y^{''}+6y^'+10y=0,y(0)=4 is y=e^{-3t}(4cos(t)+c_{2}sin(t))

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y^{''}+6y^'+10y=0,y(0)=4

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (0) = 4

y = e^{- 3 t} (4 cos (t) + c_{2} sin (t))

Solution steps

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

Solve linear ODE:

y = e^{- 3 t} (4 cos (t) + c_{2} sin (t))

y = e^{- 3 t} (4 cos (t) + c_{2} sin (t))

6 y^{^{''}} + 5 y^{^{'}} - 6 y = 0

y^{^{''}} + \frac{4}{3} y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{16}{3}

y^{^{''}} - y = 0, y (1) = 0, y^{^{'}} (1) = e

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

\frac{d ^{2} y}{d x ^{2}} - 36 \frac{d y}{d x} = 0

What is the solution for y^{''}+6y^'+10y=0,y(0)=4 ?
The solution for y^{''}+6y^'+10y=0,y(0)=4 is y=e^{-3t}(4cos(t)+c_{2}sin(t))