taylor e^{-2x}sin(x),0

Lösung

taylor

e^{- 2 x} sin (x), 0

x - 2 x^{2} + \frac{11}{6} x^{3} - x^{4} + \frac{41}{120} x^{5} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{- 4}{2 !} x^{2} + \frac{11}{3 !} x^{3} + \frac{- 24}{4 !} x^{4} + \frac{41}{5 !} x^{5} + \dots

Fasse zusammen

= x - 2 x^{2} + \frac{11}{6} x^{3} - x^{4} + \frac{41}{120} x^{5} + \dots

\int \frac{5 x ^{2} + 5 x + 3}{x ( x + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{6 x ^{2}}{( x ^{3} + 1 ) ^{2}})

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = - 3

\int \frac{1}{1 - 3 y} d y

y^{^{''}} + 2 y^{^{'}} = 8 sin (2 t) + 32 cos (2 t)