y^{''}+16y=-24sin(4t)

Lösung

y^{^{''}} + 16 y = - 24 sin (4 t)

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

Schritte zur Lösung

y^{''} (t) + 16 y = - 24 sin (4 t)

Solve linear ODE:

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

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

\int x^{2} x^{3} + 2 d x

n = 0 \sum \infty \frac{1}{4 ^{n + 1}}

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

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

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