y^'=e^{(2+4)x+(1+3)y}

Lösung

y^{^{'}} = e^{(2 + 4) x + (1 + 3) y}

y = - \frac{ln ( - \frac{2}{3} e ^{6 x} - c _{1} )}{4}

Schritte zur Lösung

y^{'} (x) = e^{(2 + 4) x + (1 + 3) y}

Solve separable ODE:

y = - \frac{ln ( - \frac{2}{3} e ^{6 x} - c _{1} )}{4}

y = - \frac{ln ( - \frac{2}{3} e ^{6 x} - c _{1} )}{4}

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

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

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

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

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