(1+y^2)dx=(x+x^2)dy

Lösung

(1 + y^{2}) d x = (x + x^{2}) d y

y = tan (ln (x) - ln (x + 1) - c_{1})

Schritte zur Lösung

(1 + y^{2}) d x = (x + x^{2}) d y

Solve separable ODE:

y = tan (ln (x) - ln (x + 1) - c_{1})

y = tan (ln (x) - ln (x + 1) - c_{1})

cos (x) d x + (1 + \frac{1}{y}) sin (x) d y = 0

(t^{3} + 7 t^{3} x^{2}) d t + x e^{t^{4}} d x = 0

y^{^{'}} = k \frac{y}{x}

\frac{d y}{d x} = - \frac{9}{6} x^{2} y^{6}

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