y^'-2(x+1)^3= 4/((x+1))y

Lösung

y^{^{'}} - 2 (x + 1)^{3} = \frac{4}{( x + 1 )} y

y = 2 ln (x + 1) (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1) + c_{1} (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1)

Schritte zur Lösung

y^{'} (x) - 2 (x + 1)^{3} = \frac{4}{( x + 1 )} y

Solve linear ODE:

y = 2 ln (x + 1) (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1) + c_{1} (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1)

y = 2 ln (x + 1) (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1) + c_{1} (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1)

\frac{d}{d x} (x^{5} x)

d er i v a t i v e f (x) = 3 x - 4

\int_{- \infty}^{1} e^{4 x} d x

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

n = 2 \sum \infty \frac{9}{n ( n + 2 )}