The integrating factor of differential equation
$\left( {1 - {x^2}} \right)\frac{{dy}}{{dx}} - xy = 1$ is
- (a) $- x$
- (b) $\frac{x}{{1 + {x^2}}}$
- (c) $\sqrt {1 - {x^2}}$ ✓ Correct
- (d) $\frac{1}{2}\log \left( {1 - {x^2}} \right)$
The integrating factor of differential equation
$\left( {1 - {x^2}} \right)\frac{{dy}}{{dx}} - xy = 1$ is
Given that,
$\left( {1 - {x^2}} \right)\frac{{dy}}{{dx}} - xy = 1$
$\Rightarrow$ $\frac{{dy}}{{dx}} - \frac{x}{{1 - {x^2}}}y = \frac{1}{{1 - {x^2}}}$
which is a linear differential equation.
$\therefore$ ${\rm{IF}} = {e^{ - \int {\frac{x}{{1 - {x^2}}}} dx}}$
Put $1 - {x^2} = t \Rightarrow - 2xdx = dt \Rightarrow xdx = - \frac{{dt}}{2}$
Now, ${\rm{IF}} = {e^{\frac{1}{2}\int {\frac{{{\rm{Jt}}}}{t}} }} = {e^{\frac{1}{2}\log t}} = {e^{\frac{1}{2}\log \left( {1 - {x^2}} \right)}} = \sqrt {1 - {x^2}}$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Differential Equations. Curated by Sachin Sharma. Free for all students.