y^'-e^yx=0

Lösung

y^{^{'}} - e^{y} x = 0

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

Schritte zur Lösung

y^{'} (x) - e^{y} x = 0

Solve separable ODE:

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

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

x \cdot \frac{d y}{d x} = \frac{1}{y ^{3}}

y^{^{'}} = x^{6} y^{- 5}

\frac{d y}{d x} = 3 x (1 + y^{2})

\frac{d p}{d t} = 2.5 p - 5 p^{2}

\frac{d y}{d x} = x e^{5 y - 3 x}, y (0) = 1