Question

Write integrating factor of the differncial equation 1-x2 dy/dx -xy=1

Answer 1

Answer:

$\sqrt{1-x^{2} }$

Step-by-step explanation:

Given Eqn.

1 - $x^{2}$ $\frac{dy}{dx}$ -xy = 1

or  $\frac{dy}{dx}$ - $(\frac{x}{1 - x^{2} })y$ = $\frac{1}{1 - x^{2} }$

or   $\frac{dy}{dx}$ + $(-\frac{x}{1 - x^{2} } )y$  = $\frac{1}{1 - x^{2} }$

Comparing above Eqn with

$\frac{dy}{dx} + Py = Q$

Here,

          $P = \frac{-x}{1 - x^{2} }$

 

          Integrating factor = I.F. = $e^{\int {P} \, dx }$

⇒ I.F. = $e^{\int {\frac{-x}{1 - x^{2} } } \, dx }$

Let I = $\int {\frac{-x}{1-x^{2} } \, dx$

Let $( 1 - x^{2} ) = t$

differentiating w.r.t x

$\frac{dt}{dx} = -2x$

or $\frac{dt}{2}$ = -x dx

⇒ I = $\int {\frac{1}{2t} } \, dt$

on integration

I = $\frac{ln(t)}{2}$

substituting t with 1 - $x^{2}$

I = $\frac{1}{2} ln(1 - x^{2} )$

or  I = $ln(1-x^{2}) ^{\frac{1}{2} }$                     {  $a(logb) = log(b^{a} )$  }

so

I.F. = $e^{ln(1-x^{2} )^{\frac{1}{2} } }$

or I.F. = $(1-x^{2} )^{\frac{1}{2}(ln(e)) }$        { $a^{ln(b)} = b^{ln(a)}$ }

or I.F. = $(1-x^{2} )^{\frac{1}{2} }$                                     { ln(e) = 1 }

or integrating factor = $\sqrt{1-x^{2} }$