y^{''}=6x+2,y(1)=-1,y^'(1)=7

Lösung

y^{^{''}} = 6 x + 2, y (1) = - 1, y^{^{'}} (1) = 7

y = x^{3} + x^{2} + 2 x - 5

Schritte zur Lösung

y^{''} (x) = 6 x + 2

Löse

y^{''} (x) = 6 x + 2 : y^{'} (x) = 3 x^{2} + 2 x + c_{1}

Löse

y^{'} (x) = 3 x^{2} + 2 x + c_{1} : y = x^{3} + x^{2} + c_{1} x + c_{2}

Wende die Anfangsbedingung an:

y = x^{3} + x^{2} + 2 x - 5

y = x^{3} + x^{2} + 2 x - 5

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

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

x \to 0 lim (\frac{1}{x - 3})

d er i v a t i v e \frac{t}{7}

\frac{d}{d x} (\frac{1}{2 x})