Discussion Forum

 The function y=f(x)  is the solution of the differential equation $\frac{dy}{dx}+\frac{xy}{x^{2}-1}=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}$   in (-1,1)

 satisfying f(0)=0, Then,   $\int_{-\frac{\sqrt{3}}{2}}^{\sqrt{3}/2} f(x) dx$  is

Answer:

Explanation:

 Plan 

 (i) solution  of the differential equation

 $\frac{dy}{dx}+Py=Q$  is 

 $Y.(IF)=\int_{}^{} Q.(IF) dx+c$

where,IF=$e^{\int P dx}$

(ii)$\int_{-a}^{a }f(x)dx=2\int_{0}^{a} f(x)dx,$

 if  f(-x)= f(x)

 Given differential  equation

$\frac{dy}{dx}+\frac{x}{x^{2}-1}y=\frac{x^{4}+2x}{\sqrt{1-x^{2}}}$

 This is a linear differential solution

 IF=   $e^{\int \frac{x}{x^{2}-1}dx}=e^{\frac{1}{2}ln|x^{2}-1|}$

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

 $\Rightarrow$   Solution  is

$y\sqrt{1-x^{2}}=\int_{}^{} \frac{x(x^{3}+2)}{\sqrt{1-x^{2}}}.\sqrt{1-x^{2}}dx$

  or    $y\sqrt{1-x^{2}}=\int_{}^{} (x^{4}+2x) dx$

$=\frac{x^{5}}{5}+x^{2}=c$

    f(0)=0 $\Rightarrow$ c=0

  $\Rightarrow$      $f(x)\sqrt{1-x^{2}}=\frac{x^{5}}{5}+x^{2}$

 Now,    $\int_{-\sqrt{3}/2}^{\sqrt{3}/2}f(x) dx= \int_{-\sqrt{3}/2}^{\sqrt{3}/2}\frac{x^{2}}{\sqrt{1-x^{2}}}dx$

                                                                                                                         [using property]

 $=2\int_{0}^{\sqrt{3}/2}\frac{x^{2}}{\sqrt{1-x^{2}}}dx$

     $=2\int_{0}^{{\pi}/3}\frac{\sin^{2}\theta}{\cos\theta}\cos\theta d\theta$

                              [ taking x= sin $\theta$]

                $=2\int_{0}^{{\pi}/3}\sin^{2}\theta d\theta$

   $=\int_{0}^{{\pi}/3}(1-cos2\theta) d\theta$

   =  $\left(\theta-\frac{\sin 2\theta}{2}\right)_{0}^{\pi/3}$

  =  $\frac{\pi}{3}-\frac{\sin 2\pi/3}{2}$

          =  $\frac{\pi}{3}-\frac{\sqrt{3}}{4}$