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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The integrating factor of the differential equation $\Large \frac{dy}{dx}\normalsize+y=\Large \frac{1+y}{x}$ is

\[(A)\;\frac{x}{e^x} \quad (B)\;\frac{e^x}{x} \quad (C)\;xe^x \quad (D)\;e^x\]

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1 Answer

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Toolbox:
  • If the linear differential equation is of the form $\large\frac{dy}{dx}$$+Py=Q,$ then its general solution is given by $ye^{\int pdx}=\int Q. e ^{\int pdx}.dx+c$
  • Here $e ^{\int pdx}$ is the integrating factor (I.F)
Given $ \large\frac{dy}{dx}$$+y=\large\frac{1+y}{x}$
This can be written as
$\large\frac{dy}{dx}$$+y=\large\frac{y}{x}+\frac{1}{x}$
$=>\large\frac{dy}{dx}$$+y\bigg(1-\large\frac{1}{x}\bigg)$$=\large\frac{1}{x}$
$=>\large\frac{dy}{dx}$$+\bigg(\large\frac{1-x}{x}\bigg)$$y=\large\frac{1}{x}$
Clearly this is a linear differential equation of the form
$\large\frac{dy}{dx}$$+Py=Q$
Where $ P=\large\frac{1-x}{x}$ and $Q=\large\frac{1}{x}$
The integrating factor is $e^{\int pdx}$
$\int pdx=\int \large\frac{1-x}{x}=\int \large \frac{1}{x} $$dx-\int dx$
On integrating we get,
$\log x-x$
Hence $ I.F=e^{\int pdx}=e^{\large[\log x-x]}$
$\Large\frac{e^{\Large\log x}}{e^x}$
But $e^{\log x}=x$
Hence $I.F=x e^{-x}$
$=\Large\frac{x}{e^x}$
Hence the correct option is $A$
answered May 21, 2013 by meena.p
 

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