taylor x^2e^{3x}

Lösung

taylor

x^{2} e^{3 x}

x^{2} + 3 x^{3} + \frac{9}{2} x^{4} + \frac{9}{2} x^{5} + \frac{27}{8} x^{6} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( x ^{2} e ^{3 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{2} e ^{3 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{2} e ^{3 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{2}{2 !} x^{2} + \frac{18}{3 !} x^{3} + \frac{108}{4 !} x^{4} + \frac{540}{5 !} x^{5} + \frac{2430}{6 !} x^{6} + \dots

Fasse zusammen

= x^{2} + 3 x^{3} + \frac{9}{2} x^{4} + \frac{9}{2} x^{5} + \frac{27}{8} x^{6} + \dots

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = b, y^{^{'}} (0) = 8

\int ln (x) x^{'} d x

\frac{d}{d x} (3 e^{x^{2} + x})

y y^{^{'}} = \frac{7}{x}

\int 2 x - \frac{1}{1 + x ^{2}} d x