y^'=(x+xy^2)/(4x)

Lösung

y^{^{'}} = \frac{x + x y ^{2}}{4 x}

y = tan (\frac{x + c _{1}}{4})

Schritte zur Lösung

y^{'} (x) = \frac{x + x y ^{2}}{4 x}

Solve separable ODE:

y = tan (\frac{x + c _{1}}{4})

y = tan (\frac{x + c _{1}}{4})

\frac{d y}{d x} = \frac{sin ( 5 x )}{y}

\frac{d}{d x} ((y) \cdot 10) + 5 y = 0

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

\frac{d v}{d t} = 8 - 7 v

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