What is the solution for y^{''''}+13y^{''}+36y=0 ?

The solution for y^{''''}+13y^{''}+36y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)+c_{3}cos(3t)+c_{4}sin(3t)

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y^{''''}+13y^{''}+36y=0

y^{^{''''}} + 13 y^{^{''}} + 36 y = 0

y = c_{1} cos (2 t) + c_{2} sin (2 t) + c_{3} cos (3 t) + c_{4} sin (3 t)

Solution steps

y^{(4)} (t) + 13 y^{''} (t) + 36 y = 0

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + c_{3} cos (3 t) + c_{4} sin (3 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + c_{3} cos (3 t) + c_{4} sin (3 t)

y = 2 x + 24, y = x^{2}

y^{^{''}} + \frac{3}{x} y^{^{'}} - \frac{15}{x ^{2}} y = 0

\int \frac{x + 1}{x - 3} d x

\int tan^{5} (6 x) sec^{4} (6 x) d x

\frac{d}{d x} (in (x sin (x)))

What is the solution for y^{''''}+13y^{''}+36y=0 ?
The solution for y^{''''}+13y^{''}+36y=0 is y=c_{1}cos(2t)+c_{2}sin(2t)+c_{3}cos(3t)+c_{4}sin(3t)