solve-xy-x-1-y-e-x-ln-1-x-2-

Question Number 90973 by abdomathmax last updated on 27/Apr/20

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

Commented by mathmax by abdo last updated on 27/Apr/20

(h e) \to {x y}^{^{'}} + (x + 1) y = 0 \Rightarrow {x y}^{^{'}} = - (x + 1) y \Rightarrow \frac{y^{^{'}}}{y} = - \frac{x + 1}{x}

= - 1 - \frac{1}{x} \Rightarrow l n ∣ y ∣ = - x - l n ∣ x ∣ + c \Rightarrow y (x) = k e^{- x} \times \frac{1}{∣ x ∣}

s o l u t i o n o n] 0, + \infty [\Rightarrow y (x) = k \frac{e^{- x}}{x} l e t u s e m v c m e t h o d

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

(e) \Rightarrow k^{^{'}} e^{- x} - \frac{k}{x} e^{- x} - {k e}^{- x} + k (x + 1) \frac{e^{- x}}{x} = e^{- x} l n (1 + x^{2}) \Rightarrow

k^{^{'}} e^{- x} - \frac{k}{x} e^{- x} - k e^{- x} + {k e}^{- x} + \frac{k}{x} e^{- x} = e^{- x} l n (1 + x^{2}) \Rightarrow

k^{^{'}} = l n (1 + x^{2}) \Rightarrow k (x) = \int l n (1 + x^{2}) d x + c

b u t \int l n (1 + x^{2}) d x =_{b y p a r t s} x l n (1 + x^{2}) - \int x \frac{2 x}{1 + x^{2}} d x

= x l n (1 + x^{2}) - 2 \int \frac{x^{2} + 1 - 1}{1 + x^{2}} d x

= x l n (1 + x^{2}) - 2 + 2 a r c t a n (x) \Rightarrow

k (x) = x l n (1 + x^{2}) + 2 a r c t a n (x) + C \Rightarrow

y (x) = \frac{e^{- x}}{x} k (x) = \frac{e^{- x}}{x} (x l n (1 + x^{2}) + 2 a r c t a n (x) + C)

Commented by mathmax by abdo last updated on 27/Apr/20

f o r g i v e \int l n (1 + x^{2}) d x = x l n (1 + x^{2}) - 2 x + 2 a r c t a n (x) + c \Rightarrow

y (x) = \frac{e^{- x}}{x} (x l n (1 + x^{2}) - 2 x + 2 a r c t a n (x) + C)

Answered by Joel578 last updated on 27/Apr/20

y^{'} (x) + p (x) y (x) = r (x)

Has the solution

y (x) = \frac{1}{F (x)} [\int F (x) r (x) d x + C]

where F (x) is integrating factor and C is integration constant

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

Integrating factor

F (x) = \exp [\int 1 + \frac{1}{x} d x] = \exp (x + \ln x) = {x e}^{x}

hence

y (x) = \frac{e^{- x}}{x} [\int {x e}^{x} . \frac{e^{- x} \ln (1 + x^{2})}{x} d x + C]

= \frac{e^{- x}}{x} [\int \ln (1 + x^{2}) d x + C]

Use integration by part : u = \ln (1 + x^{2}), d v = d x

I = x \ln (1 + x^{2}) - \int \frac{2 x^{2}}{1 + x^{2}} d x

= x \ln (1 + x^{2}) + \int \frac{2 - 2 - 2 x^{2}}{1 + x^{2}} d x

= x \ln (1 + x^{2}) + \int \frac{2}{1 + x^{2}} - 2 d x

= x \ln (1 + x^{2}) + {2 \tan}^{- 1} x - 2 x

∴ y (x) = \frac{e^{- x}}{x} [x \ln (1 + x^{2}) + {2 \tan}^{- 1} x - 2 x + C]

Commented by mathmax by abdo last updated on 27/Apr/20

t h a n k y o u s i r .