16y^{''}+7y=0

Lösung

16 y^{^{''}} + 7 y = 0

y = c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4})

Schritte zur Lösung

16 y^{''} (t) + 7 y = 0

Solve linear ODE:

y = c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4})

y = c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4})

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

\frac{\partial}{\partial x} (sin (ln (x)))

\frac{\partial}{\partial y} (x^{- 7} y^{2} + x y^{2} + 3 x y)

\frac{d y}{d x} = - 10 y, y (0) = 5

\int_{0}^{\frac{1}{2}} 8 x (\frac{1}{2} - x) d x