y^{''}+25y=4csc^2(5t)

Lösung

y^{^{''}} + 25 y = 4 csc^{2} (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{- 4 ln ( tan ( \frac{5 t}{2} ) ) cos ( 5 t ) - 4}{25}

Schritte zur Lösung

y^{''} (t) + 25 y = 4 csc^{2} (5 t)

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{- 4 ln ( tan ( \frac{5 t}{2} ) ) cos ( 5 t ) - 4}{25}

y = c_{1} cos (5 t) + c_{2} sin (5 t) + \frac{- 4 ln ( tan ( \frac{5 t}{2} ) ) cos ( 5 t ) - 4}{25}

\frac{\partial}{\partial x} (3 x cos (y))

x \to 0 lim (\frac{1}{cos ( 8 x )})

\frac{d}{d x} (- 11 (4 x^{3} - 11)^{13})

t an g e n t \frac{6 x}{x + 3}

\frac{\partial}{\partial x} (u v^{2})