What is the solution for y^{''}-8y^'+41y=0 ?

The solution for y^{''}-8y^'+41y=0 is y=e^{4t}(c_{1}cos(5t)+c_{2}sin(5t))

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y^{''}-8y^'+41y=0

y^{^{''}} - 8 y^{^{'}} + 41 y = 0

y = e^{4 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) - 8 y^{'} (t) + 41 y = 0

Solve linear ODE:

y = e^{4 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{4 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

\frac{d}{d x} ((x^{2} - x + 1)^{- 7})

\int \frac{x - 1}{x ( x ^{2} - 5 x + 6 )} d x

\int (3 - 5 x)^{2} d x

\frac{d y}{d x} = 2 y cos (x)

x \to 4 lim (- x^{2} - 9 x - 8)

What is the solution for y^{''}-8y^'+41y=0 ?
The solution for y^{''}-8y^'+41y=0 is y=e^{4t}(c_{1}cos(5t)+c_{2}sin(5t))