(2x^2y+2y+5)dx+(2x^3+2x)dy=0

Lösung

(2 x^{2} y + 2 y + 5) d x + (2 x^{3} + 2 x) d y = 0

y = - \frac{5 x arctan ( x )}{2 ( x ^{2} + 1 )} - \frac{c _{1} x}{2 ( x ^{2} + 1 )} - \frac{c _{1}}{2 x ( x ^{2} + 1 )} - \frac{5 arctan ( x )}{2 x ( x ^{2} + 1 )}

Schritte zur Lösung

(2 x^{2} y + 2 y + 5) d x + (2 x^{3} + 2 x) d y = 0

Solve using substitution:

y = - \frac{5 x arctan ( x )}{2 ( x ^{2} + 1 )} - \frac{c _{1} x}{2 ( x ^{2} + 1 )} - \frac{c _{1}}{2 x ( x ^{2} + 1 )} - \frac{5 arctan ( x )}{2 x ( x ^{2} + 1 )}

y = - \frac{5 x arctan ( x )}{2 ( x ^{2} + 1 )} - \frac{c _{1} x}{2 ( x ^{2} + 1 )} - \frac{c _{1}}{2 x ( x ^{2} + 1 )} - \frac{5 arctan ( x )}{2 x ( x ^{2} + 1 )}

\int e^{- a x} x d x

d er i v a t i v e f (x) = \frac{3 x}{7 x ^{2} + 1}

\frac{\partial}{\partial x} (3 y - \frac{x}{y ^{2}})

a re a tan (3 x), 2 sin (3 x), - \frac{π}{9}, \frac{π}{9}

d er i v a t i v e 4 sec^{2} (x)