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

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Toolbox:
• 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$(1-x^2)$
$\large\frac{dy}{dx}-\frac{x}{(1-x^2)}$$y=\large\frac{1}{1-x^2} Clearly this is of the form \large\frac{dy}{dx}$$+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,

1 answer

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