What is the solution for y^{''}-6y^'+34y=0 ?

The solution for y^{''}-6y^'+34y=0 is y=e^{3t}(c_{1}cos(5t)+c_{2}sin(5t))

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y^{''}-6y^'+34y=0

y^{^{''}} - 6 y^{^{'}} + 34 y = 0

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 34 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (\frac{1}{3} x^{3})

\int_{- 1}^{2} (7 - 3 t) d t

\int csc (d) x d x

(- 7, - 3), (- 5, - 8)

\int sin^{4} (θ) cos^{2} (θ) d θ

What is the solution for y^{''}-6y^'+34y=0 ?
The solution for y^{''}-6y^'+34y=0 is y=e^{3t}(c_{1}cos(5t)+c_{2}sin(5t))