What is the solution for x^{''}+6x^'+34x=68 ?

The solution for x^{''}+6x^'+34x=68 is x=e^{-3t}(c_{1}cos(5t)+c_{2}sin(5t))+2

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x^{''}+6x^'+34x=68

x^{^{''}} + 6 x^{^{'}} + 34 x = 68

x = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t)) + 2

Solution steps

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

Solve linear ODE:

x = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t)) + 2

x = e^{- 3 t} (c_{1} cos (5 t) + c_{2} sin (5 t)) + 2

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 x^{2} - 1

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} = - 3

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

x^{^{''}} + 4 x = 12 cos (t)

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = x^{2}

What is the solution for x^{''}+6x^'+34x=68 ?
The solution for x^{''}+6x^'+34x=68 is x=e^{-3t}(c_{1}cos(5t)+c_{2}sin(5t))+2