y^'=(x+y)^2-1

Lösung

y^{^{'}} = (x + y)^{2} - 1

y = - \frac{1}{x + c _{1}} - x

Schritte zur Lösung

y^{'} (x) = (x + y)^{2} - 1

Solve using substitution:

y = - \frac{1}{x + c _{1}} - x

y = - \frac{1}{x + c _{1}} - x

tan (f (x)) (x) = - 4 x + \frac{1}{2} x - 5, at x = 1

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

d er i v a t i v e s (t) = - 9.8 t^{2} + 2 t + 1

d er i v a t i v e y = (2 x + 3)^{5} + 10 x (2 x + 3)^{4}

\int e x^{2} + π d x