y^'=0.4(10-x)(5-x)

Lösung

y^{^{'}} = 0.4 (10 - x) (5 - x)

y = 0.4 (50 x - 7.5 x^{2} + 0.33333 \dots x^{3}) + c_{1}

Schritte zur Lösung

y^{'} (x) = 0.4 (10 - x) (5 - x)

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

y = \int 0.4 (10 - x) (5 - x) d x

\int 0.4 (10 - x) (5 - x) d x = 0.4 (50 x - 7.5 x^{2} + 0.33333 \dots x^{3}) + c_{1}

y = 0.4 (50 x - 7.5 x^{2} + 0.33333 \dots x^{3}) + c_{1}