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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{1}{8} u^{^{''}} + \frac{4}{5} u^{^{'}} + 28 H (t) = 0

y^{^{''}} + 10 y^{^{'}} + 15 y = 0

y^{^{''}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 1

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

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

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