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

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

x y^{^{'}} - y = 3

(5 + y) y^{^{'}} = sin (7 + x)

y^{^{'}} + 6 e^{x + y} = 0

(1 - cos (x)) d y + (2 y sin (x) tan (x)) d x = 0

\frac{d y}{d t} - y = 0, y (0) = 10