e^yy^'=e^y+4x

Lösung

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

y = ln (c_{1} + 4 (- e^{- x} x - e^{- x})) + x

Schritte zur Lösung

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

Solve exact ODE:

y = ln (c_{1} + 4 (- e^{- x} x - e^{- x})) + x

y = ln (c_{1} + 4 (- e^{- x} x - e^{- x})) + x

d er i v a t i v e sec^{2} (2 x)

y d x + (x^{2} + 4 x) d y = 0

\frac{d y}{d x} = (e^{x} + 7)^{2}

d er i v a t i v e y = x^{2} cos (\frac{4}{x})

n or ma l f (x) = x^{4} + 2 e^{x}, (0, 2)