y^'=((e^{2x+y}))/y

Lösung

y^{^{'}} = \frac{( e ^{2 x + y} )}{y}

y = - W (\frac{e ^{2 x} + c _{1}}{2 e}) - 1

Schritte zur Lösung

y^{'} (x) = \frac{( e ^{2 x + y} )}{y}

Solve separable ODE:

y = - W (\frac{e ^{2 x} + c _{1}}{2 e}) - 1

y = - W (\frac{e ^{2 x} + c _{1}}{2 e}) - 1

(5 + x) y^{^{'}} = 2 y

(1 - y^{2}) d x = - (x y - y) d y

e^{- y} + e^{- y} y^{^{'}} = 1

y^{^{'}} = - \frac{1}{y}

[1 + t^{2}] \frac{d y}{d t} = 1 + y^{2}