y^'=sin(5x),y(0)=2

Lösung

y^{^{'}} = sin (5 x), y (0) = 2

y = - \frac{1}{5} cos (5 x) + \frac{11}{5}

Schritte zur Lösung

y^{'} (x) = sin (5 x)

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

y = \int sin (5 x) d x

\int sin (5 x) d x = - \frac{1}{5} cos (5 x) + c_{1}

y = - \frac{1}{5} cos (5 x) + c_{1}

Wende die Anfangsbedingung an:

y = - \frac{1}{5} cos (5 x) + \frac{11}{5}

y = - \frac{1}{5} cos (5 x) + \frac{11}{5}

(x + 1) \frac{d y}{d x} + (x + 2) y = 8 x e^{- x}

\frac{\partial}{\partial x} (5 x^{3} y z^{2})

\int 3 x^{4} d x

\int_{- 1}^{2} x^{3} - 1 d x

x \to 0 lim (x^{4})