xy+xy^'=x

Lösung

x y + x y^{^{'}} = x

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

\frac{d x}{d t} = t x^{2}

\frac{d y}{e ^{2 y}} = e^{3 x} d x

y^{2} sec (7 x) \frac{d y}{d x} = 1, y (0) = 0

5 x y + y^{^{'}} x = 0

\frac{d y}{d x} = y (x - 2)