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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{1}{x} + \frac{3}{x ^{2}} + \frac{2}{x ^{3}})

x \to \infty lim (co d s)

\int_{1}^{e} x (ln (x))^{2} d x

\frac{d}{d x} (cos (e^{x^{4} s i n (x)}))

\int \frac{x ^{6} - 3 x}{x ^{4}} d x

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