y^'+(2x-1)y^2=0

Lösung

y^{^{'}} + (2 x - 1) y^{2} = 0

y = - \frac{1}{- x ^{2} + x + c _{1}}

Schritte zur Lösung

y^{'} (x) + (2 x - 1) y^{2} = 0

Solve separable ODE:

y = - \frac{1}{- x ^{2} + x + c _{1}}

y = - \frac{1}{- x ^{2} + x + c _{1}}

\frac{d Q}{d t} = 4 Q^{2} ln (t)

x d x + y e^{y} d y = 0

2 x y d x - (1 - x^{2}) d y = 0

2 (x - 1) y^{^{'}} = 3 y

d y = (1 + 4 y^{2}) d x