(7x+8)+(7y-7)y^'=0

Lösung

(7 x + 8) + (7 y - 7) y^{^{'}} = 0

y = \frac{14 + 196 - 28 ( - c _{1} + 16 x + 7 x ^{2} )}{14}, y = \frac{14 - 196 - 28 ( - c _{1} + 16 x + 7 x ^{2} )}{14}

Schritte zur Lösung

(7 x + 8) + (7 y - 7) y^{'} (x) = 0

Solve separable ODE:

y = \frac{14 + 196 - 28 ( - c _{1} + 16 x + 7 x ^{2} )}{14}, y = \frac{14 - 196 - 28 ( - c _{1} + 16 x + 7 x ^{2} )}{14}

y = \frac{14 + 196 - 28 ( - c _{1} + 16 x + 7 x ^{2} )}{14}, y = \frac{14 - 196 - 28 ( - c _{1} + 16 x + 7 x ^{2} )}{14}

\int \frac{1}{tan ^{2} ( x )} d x

d er i v a t i v e x^{11} arccos (x)

t an g e n t g (x) = e^{x^{5}}, (- 1, \frac{1}{e})

\frac{\partial}{\partial x} (2 ln (x))

b \to \infty lim (b c)