y^'=e^{9x}-4x

Lösung

y^{^{'}} = e^{9 x} - 4 x

y = \frac{1}{9} e^{9 x} - 2 x^{2} + c_{1}

Schritte zur Lösung

y^{'} (x) = e^{9 x} - 4 x

Wenn

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

dann

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

y = \int e^{9 x} - 4 x d x

\int e^{9 x} - 4 x d x = \frac{1}{9} e^{9 x} - 2 x^{2} + c_{1}

y = \frac{1}{9} e^{9 x} - 2 x^{2} + c_{1}

\int e^{- 2 x^{2}} \cdot (- 4) x^{3} d x

n = 1 \sum \infty (1 + \frac{3}{n})^{n}

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

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

\int_{1}^{9} u + 2 d u