What is the solution for x^{''}-10x^'+29x=0 ?

The solution for x^{''}-10x^'+29x=0 is x=e^{5t}(c_{1}cos(2t)+c_{2}sin(2t))

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x^{''}-10x^'+29x=0

x^{^{''}} - 10 x^{^{'}} + 29 x = 0

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} + 3 y^{^{''}} - 24 y^{^{'}} + 28 y = 0

y^{^{''}} - 15 y^{^{'}} - 56 y = 0

4 y^{^{''}} - 21 y^{^{'}} + 20 y = 0, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 5 y = 0

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

What is the solution for x^{''}-10x^'+29x=0 ?
The solution for x^{''}-10x^'+29x=0 is x=e^{5t}(c_{1}cos(2t)+c_{2}sin(2t))