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Integrating factor of the differential equation $(1-x^2)\Large \frac{dy}{dx}\normalsize -xy=1$ is \[(A)\;-x\quad(B)\;\frac{x}{1-x^2}\quad(C)\;\sqrt {1-x^2}\quad(D)\;\frac{1}{2}log(1-x^2)\]

1 Answer

  • A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$ has the general solution $y e^{\large\int pdx}=\int Q.e^{\large \int pdx}.dx+c$. where $e^{\large \int pdx}$ is the integrating factor (I.F)
Given $(1-x^2)\large\frac{dy}{dx}$$-xy=1$
Divide throughout by$(1-x^2)$
Clearly this is of the form $\large\frac{dy}{dx}$$+Py=Q$ which repreaents a linear differential equation.
Here $P= \large\frac{-x}{1-x^2}$ and $ Q=\large\frac{1}{1-x^2}$
The integrating factor (I.F) is $e^{\int pdx}$
$\int pdx= \int \large\frac{-x}{1-x^2}$$dx$
Put $1-x^2=t;$ on differentiating w.r.t x we get,
$-2xdx=dt \; or \; -x dx=dt/2$
Therefore $ \int pdx=\int \large\frac{dt/2}{t}$
$=\large\frac{1}{2} $$\log |t|$
Substituting for t we get,
$ \int pdx=\large\frac{1}{2} $$\log (1-x^2)$
$=\log (1-x^2)^{1/2}$
Hence I.F is $ e^{\large\log (1-x^2)^{1/2}}$
But $ e^{\large\log (1-x^2)^{1/2}}=(1-x^2)^{1/2}=\sqrt {1-x^2}$
Hence $C$ is the correct option
answered May 16, 2013 by meena.p

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