solve-1-e-x-y-y-e-x-1-x-2-

Question Number 90968 by abdomathmax last updated on 27/Apr/20

s o l v e (1 + e^{x}) y^{^{'}} - y = \frac{e^{x}}{1 + x^{2}}

Commented by niroj last updated on 27/Apr/20

(1 + e^{x}) y^{^{'}} - y = \frac{e^{x}}{1 + x^{2}}

\frac{dy}{dx} - \frac{1}{1 + e^{x}} y = \frac{e^{x}}{(1 + e^{x}) (1 + x^{2})}

P = - \frac{1}{1 + e^{x}}, Q = \frac{e^{x}}{(1 + e^{x}) (1 + x^{2})}

IF = e^{\int Pdx} = e^{- \int \frac{1}{1 + e^{x}} dx}

- \int \frac{1}{e^{x} (e^{- x} + 1)} dx

- \int \frac{e^{- x} dx}{e^{- x} + 1}

put e^{- x} + 1 = t

- e^{- x} dx = dt

\int \frac{1}{t} dt = logt = \log (e^{- x} + 1)

IF = e^{\log (e^{- x} + 1)} = e^{- x} + 1

we know,

y . IF = \int IF . Qdx + C

y . (e^{- x} + 1) = \int (e^{- x} + 1) . \frac{e^{x}}{(1 + e^{x}) (1 + x^{2})} dx + C

y (\frac{1}{e^{x}} + 1) = \int \frac{(1 + e^{x})}{e^{x}} . \frac{e^{x}}{(1 + e^{x}) (1 + x^{2})} dx + C

\frac{y (1 + e^{x})}{e^{x}} = \int \frac{1}{1 + x^{2}} dx + C

y = \frac{e^{x} (\tan^{- 1} x + C)}{(1 + e^{x})} / / .

Commented by mathmax by abdo last updated on 28/Apr/20

(h e) \to (1 + e^{x}) y^{^{'}} - y = 0 \Rightarrow \frac{y^{^{'}}}{y} = \frac{1}{1 + e^{x}} \Rightarrow l n ∣ y ∣= \int \frac{d x}{1 + e^{x}} + c

=_{e^{x} = t} \int \frac{d t}{t (1 + t)} = \int (\frac{1}{t} - \frac{1}{1 + t}) d t + c = l n ∣ \frac{t}{1 + t} ∣ + c = l n ∣ \frac{e^{x}}{1 + e^{x}} ∣ + c

y (x) = k \times \frac{e^{x}}{1 + e^{x}} l e t u s e m v c m e t h o d \to y^{^{'}} = k^{^{'}} \frac{e^{x}}{1 + e^{x}}

+ k \times \frac{e^{x}}{{(1 + e^{x})}^{2}}

(e) \Rightarrow k^{^{'}} e^{x} + \frac{{k e}^{x}}{1 + e^{x}} - \frac{{k e}^{x}}{1 + e^{x}} = \frac{e^{x}}{1 + x^{2}} \Rightarrow k^{^{'}} = \frac{1}{1 + x^{2}} \Rightarrow k (x) = a r c t a n x + c

t h e g e n e r a l s o l u t i o n i s

y (x) = \frac{e^{x}}{1 + e^{x}} k (x) = \frac{e^{x}}{1 + e^{x}} (a r c t a n x + c)