y^'= 4/(ye^{x+4y)}

Lösung

y^{^{'}} = \frac{4}{y e ^{x + 4 y}}

y = \frac{W ( \frac{- 64 e ^{- x} + c _{1}}{e} )}{4} + \frac{1}{4}

Schritte zur Lösung

y^{'} (x) = \frac{4}{y e ^{x + 4 y}}

Solve separable ODE:

y = \frac{W ( \frac{- 64 e ^{- x} + c _{1}}{e} )}{4} + \frac{1}{4}

y = \frac{W ( \frac{- 64 e ^{- x} + c _{1}}{e} )}{4} + \frac{1}{4}

e^{y} y^{^{'}} = 9 x^{7} + 2

(y^{2} + 6) d x = y sec^{2} (x) d y

\frac{d y}{d x} = \frac{4 x}{y - 1}, y (2) = 3

4 y^{^{'}} - 4 y - 72 = 3 y^{^{'}}

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