What is the solution for y^{''}-4y^'+26y=0 ?

The solution for y^{''}-4y^'+26y=0 is y=e^{2t}(c_{1}cos(sqrt(22)t)+c_{2}sin(sqrt(22)t))

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y^{''}-4y^'+26y=0

y^{^{''}} - 4 y^{^{'}} + 26 y = 0

y = e^{2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 26 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

y = e^{2 t} (c_{1} cos (22 t) + c_{2} sin (22 t))

(1 - 4 y + x) \frac{d y}{d x} + y = 4 x - 1

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What is the solution for y^{''}-4y^'+26y=0 ?
The solution for y^{''}-4y^'+26y=0 is y=e^{2t}(c_{1}cos(sqrt(22)t)+c_{2}sin(sqrt(22)t))