What is the solution for x^{''}+10x^'+29x=0,x^'(0)=6 ?

The solution for x^{''}+10x^'+29x=0,x^'(0)=6 is x=e^{-5t}(-(cos(2t)(6-2c_{2}))/5+c_{2}sin(2t))

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x^{''}+10x^'+29x=0,x^'(0)=6

x^{^{''}} + 10 x^{^{'}} + 29 x = 0, x^{^{'}} (0) = 6

x = e^{- 5 t} (- \frac{cos ( 2 t ) ( 6 - 2 c _{2} )}{5} + c_{2} sin (2 t))

Solution steps

x^{''} (t) + 10 x^{'} (t) + 29 x = 0

Solve linear ODE:

x = e^{- 5 t} (- \frac{cos ( 2 t ) ( 6 - 2 c _{2} )}{5} + c_{2} sin (2 t))

x = e^{- 5 t} (- \frac{cos ( 2 t ) ( 6 - 2 c _{2} )}{5} + c_{2} sin (2 t))

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

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 10, y^{^{'}} (0) = - 9

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

y^{^{''}} - 4 y^{^{'}} - 4 y = 0, y (0) = 2, y^{^{'}} (0) = 16

What is the solution for x^{''}+10x^'+29x=0,x^'(0)=6 ?
The solution for x^{''}+10x^'+29x=0,x^'(0)=6 is x=e^{-5t}(-(cos(2t)(6-2c_{2}))/5+c_{2}sin(2t))