if-dy-dx-y-1-y-x-then-the-integrating-factor-I-F-is-

Question Number 78999 by gopikrishnan last updated on 22/Jan/20

i f d y / d x + y = 1 + y / x, t h e n t h e i n t e g r a t i n g f a c t o r (I, F) i s

Commented by mr W last updated on 22/Jan/20

p l e a s e {d o n}^{'} t t y p e t h e w h o l e p o s t i n a

s i n g l e l i n e s i r!

w i t h 1 + y / x d o y o u m e a n 1 + \frac{y}{x} o r

\frac{1 + y}{x} ?

Commented by jagoll last updated on 22/Jan/20

not clear sir your question

Answered by mind is power last updated on 22/Jan/20

i f \frac{d y}{d x} + y = 1 + \frac{y}{x}, x \in R^{*}

\Leftrightarrow \frac{d y}{d x} + y (1 - \frac{1}{x}) = 1 \dots W

H o m g e n i u s

\Rightarrow \frac{d y}{d x} + y (1 - \frac{1}{x}) = 0

\Leftrightarrow \frac{d y}{y} = (\frac{1}{x} - 1) d x, x

\Rightarrow l n (y) = (l n (x) - x)

\Rightarrow y = {k x e}^{- x}, k C o n s t a n t e

G e n e r a l y_{p} (x) = k (x) {x e}^{- x}, S o l u t i o n o f . . W

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

\Rightarrow k^{'} (x) = \frac{e^{- x}}{x} \Rightarrow k (x) = \int \frac{e^{- x}}{x} = - E_{i} (x), E x p o n e n t i e l I n t e g r a l

Y_{p} = (- E_{i} (x)) {x e}^{- x}

Y = Y_{h} + Y_{p} = {k x e}^{- x} - E_{i} (x) {x e}^{- x}, k \in R, x \in R_{+}^{*}

\frac{d y}{d x} + y = \frac{1 + y}{x}

\Leftrightarrow x \frac{d y}{d x} = y (1 - x) + 1

H o m g e n i u s

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

l e t Y_{p} = k (x) {x e}^{- x} p a r t i c u l a r S o l u t i o n \Rightarrow

x (k^{'} {x e}^{- x} + k (e^{- x} - x)) + (x - 1) {k x e}^{- x} = 1

\Rightarrow k^{'} = \frac{e^{- x}}{x^{2}} \Rightarrow k = \int \frac{e^{- x}}{x^{2}}, b y P a r t = (- \frac{e^{- x}}{x} - \int \frac{e^{- x}}{x} d x

= - \frac{e^{- x}}{x} - E_{i} (x)

Y_{p} = (- \frac{e^{- x}}{x} - E_{i} (x)) {x e}^{- x} + {k x e}^{- x}, k \in R