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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (2 x y^{2} - 2 x^{3} y)

n = 0 \sum \infty \frac{( ( 1 ) ^{n} )}{n}

d er i v a t i v e y = \frac{x ^{2} + 1}{2 x ^{2} - 1}

\frac{d}{d x} (1 - cos (x - 1))

t an g e n t f (x) = 12 x - 2 x, at x = 9