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The integrating factor of $\large\frac{dy}{dx}-y=\frac{1-y}{x}$ is__________.

1 Answer

  • A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$ has general solution as $ye^{\int pdx}=\int Q.e^{\int pdx}dx+c$
  • Where $ e^{\int pdx}$ is the integral factor (I.F)
Given $ \large\frac{dy}{dx}$$-y=\large\frac{1-y}{x}$
This can be written as
or $\large\frac{dy}{dx}$$-y(1-\large\frac{1}{x})=\frac{1}{x}$
Clearly this is of the form $\large\frac{dy}{dx}$$+Py=Q$
Where $P=\large\frac{x-1}{x} $$\qquad \bigg(1-\large\frac{1}{x}\bigg)$
$\int pdx=\int \bigg(1-\large\frac{1}{x}\bigg)$$dx$
$=(x-\log x)$
$e^{\large \int pdx}=e^{x-\log x}=e^x.e^{-\log x}$
Hence the integrating factor $ \large\frac{dy}{dx}$$-y=\large\frac{1-y}{x}$ is $\Large\frac{e^x}{x}$
answered May 22, 2013 by meena.p