What is the solution for y^{''}-4y^'+29y=0 ?

The solution for y^{''}-4y^'+29y=0 is y=e^{2t}(c_{1}cos(5t)+c_{2}sin(5t))

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y^{''}-4y^'+29y=0

y^{^{''}} - 4 y^{^{'}} + 29 y = 0

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 29 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{2 e ^{x} - x ^{5}}{1 - 3 x ^{5}})

\frac{d}{d x} (x e^{x y + 1})

f (x) = x^{2} - 3 x + 2, at x = 2

\int \frac{2 x - 3}{6} d x

ln (x^{2})

What is the solution for y^{''}-4y^'+29y=0 ?
The solution for y^{''}-4y^'+29y=0 is y=e^{2t}(c_{1}cos(5t)+c_{2}sin(5t))