y^{''}=xe^x

Lösung

y^{^{''}} = x e^{x}

y = e^{x} x - 2 e^{x} + c_{1} x + c_{2}

Schritte zur Lösung

y^{''} (x) = x e^{x}

Löse

y^{''} (x) = x e^{x} : y^{'} (x) = e^{x} x - e^{x} + c_{1}

Löse

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

y = e^{x} x - 2 e^{x} + c_{1} x + c_{2}

t an g e n t f (x) = x^{3} - 2 x^{2} + 4 x + 3, at x = 1

10 x - 8 y \cdot x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 4

d er i v a t i v e f (x) = \frac{( x - 5 )}{( x + 3 ) ^{4}}

\int_{0}^{1} - 12 e^{x} d x

\frac{d}{d x} (x^{6} e^{- x})