y^'=xe^y

Lösung

y^{^{'}} = x e^{y}

y = - ln (- \frac{x ^{2}}{2} - c_{1})

Schritte zur Lösung

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

Solve separable ODE:

y = - ln (- \frac{x ^{2}}{2} - c_{1})

y = - ln (- \frac{x ^{2}}{2} - c_{1})

\frac{\partial}{\partial y} (sin^{2} (y))

\frac{\partial}{\partial x} (x^{e^{x}})

s l o p e (- 4) (12.2)

\frac{\partial}{\partial x} ((x^{9}) (y^{6}) + 7 (x^{5}) y)

\int \frac{2 x ^{2} - 4 x + 4}{3 + 2 x - x ^{2}} d x