y^'=e^{y+8x}

Lösung

y^{^{'}} = e^{y + 8 x}

y = - ln (- \frac{1}{8} e^{8 x} - c_{1})

Schritte zur Lösung

y^{'} (x) = e^{y + 8 x}

Solve separable ODE:

y = - ln (- \frac{1}{8} e^{8 x} - c_{1})

y = - ln (- \frac{1}{8} e^{8 x} - c_{1})

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

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

x^{^{'}} = 3 x^{\frac{2}{3}}

\frac{d y}{d x} = \frac{2 x ^{4}}{y}

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