What is the solution for x^{''}+6x^'+34x=68,x(0)=5,x^'(0)=7 ?

The solution for x^{''}+6x^'+34x=68,x(0)=5,x^'(0)=7 is x=e^{-3t}(3cos(5t)+16/5 sin(5t))+2

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x^{''}+6x^'+34x=68,x(0)=5,x^'(0)=7

x^{^{''}} + 6 x^{^{'}} + 34 x = 68, x (0) = 5, x^{^{'}} (0) = 7

x = e^{- 3 t} (3 cos (5 t) + \frac{16}{5} sin (5 t)) + 2

Solution steps

x^{''} (t) + 6 x^{'} (t) + 34 x = 68

Solve linear ODE:

x = e^{- 3 t} (3 cos (5 t) + \frac{16}{5} sin (5 t)) + 2

x = e^{- 3 t} (3 cos (5 t) + \frac{16}{5} sin (5 t)) + 2

y^{^{''}} + 9 y = 13 sec^{2} (3 t)

y^{^{''}} + 2 y^{^{'}} = 2 x + 5 - e - 2 x

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 8 x

y^{^{''}} - 9 y = (x^{2} - 8) sin (3 x)

y^{^{''}} - 3 y^{^{'}} = sin (4 t)

What is the solution for x^{''}+6x^'+34x=68,x(0)=5,x^'(0)=7 ?
The solution for x^{''}+6x^'+34x=68,x(0)=5,x^'(0)=7 is x=e^{-3t}(3cos(5t)+16/5 sin(5t))+2