solve differential equation $$(x^4+x^2)y{''}-(x^3+2x)y^{'}+2y=0$$

 $Q$  $$(x^4+x^2)y{''}-(x^3+2x)y^{'}+2y=0$$

$solution:$    $we$ $have$ $given$   $$(x^4+x^2)y{''}-(x^3+2x)y^{'}+2y=0...............(i)$$

$dividing$ $both$ $sides$ $by$ $(x^4+x^2),$ $then$ $we$ $have$

$$y{''}-\frac{(x^3+2x)}{(x^4+x^2)}y^{'}+2\frac{y}{(x^4+x^2)}=0$$

$now$ $comparing$ $above$ $equation$ $with$ $\frac{d^2y}{dx^2}+P\frac{dy}{dx}+Qy=R$ 

$$P=-\frac{(x^3+2x)}{(x^4+x^2)}$$

$$Q=\frac{2}{(x^4+x^2)}$$

$$R=0$$

$we$ $see$ $that$ $2+2Px+Qx^2=2+2x\{-\frac{(x^3+2x)}{(x^4+x^2)}\}+x^2\{\frac{2}{(x^4+x^2)}\}$

$2+2Px+Qx^2= \frac{2(x^4+x^2)-2x(x^3+2x)+x^2(2)}{(x^4+x^2)}$

$2+2Px+Qx^2= \frac{2x^4+2x^2-2x^4-4x^2+2x^2}{(x^4+x^2)}$

$2+2Px+Qx^2= \frac{2x^4-2x^4-4x^2+4x^2}{(x^4+x^2)}$

$2+2Px+Qx^2=0$

$thus$ $y_1=x^2$ $is$ $a$ $part$ $of$ $complementary$ $function.$

$Now$ $putting$ $y=vy_1=vx^2$ $in$ $the$ $equation$ $(i)$

$then$ $we$ $have$ $\frac{d^2v}{dx^2}+(P+\frac{2}{y_1}\frac{dy_1}{dx})\frac{dv}{dx}=\frac{R}{y_1}$ 

$$\frac{d^2v}{dx^2}+(\frac{-x^3-2x}{x^4+x^2}+\frac{2}{x^2}2x)\frac{dv}{dx}=0$$

$$\frac{d^2v}{dx^2}+(\frac{-x^3-2x}{x^4+x^2}+\frac{4}{x})\frac{dv}{dx}=0$$

$now$ $putting$ $p=\frac{dv}{dx}$

$$\frac{dp}{dx}+(\frac{-x^3-2x}{x^4+x^2}+\frac{4}{x})p=0$$

$$\frac{dp}{dx}=(\frac{x^3+2x}{x^4+x^2}-\frac{4}{x})p$$

$$\frac{dp}{p}=(\frac{x^3+2x}{x^4+x^2}-\frac{4}{x})dx$$ 

$$\frac{dp}{p}=(\frac{4x^3+2x}{x^4+x^2}-\frac{3x^3}{x^4+x^2}-\frac{4}{x})dx$$

$$\frac{dp}{p}=(\frac{4x^3+2x}{x^4+x^2}-\frac{3x}{x^2+1}-\frac{4}{x})dx$$

$now$ $integrating$ $both$ $sides$ $after$ $putting$ $(x^4+x^2)=w$ & $(x^2+1)=z$

$$logp=log(x^4+x^2)-\frac{3}{2}log(x^2+1)-4logx+logc_1$$

$$logp=log(x^4+x^2)-log(x^2+1)^{\frac{3}{2}}-logx^4+logc_1$$

$$logp=log\frac{c_1(x^4+x^2)}{x^4(x^2+1)^{\frac{3}{2}}}$$

$$p=\frac{c_1x^2(x^2+1)}{x^4(x^2+1)\sqrt{x^2+1}}$$

$$p=\frac{c_1}{x^2\sqrt{x^2+1}}$$

$now$ $putting$ $p=\frac{dv}{dx}$

$$\frac{dv}{dx}=\frac{c_1}{x^2\sqrt{x^2+1}}$$

$$dv=\frac{c_1}{x^2\sqrt{x^2+1}}dx$$

$now$ $integrating$ $both$ $sides$ $after$ $putting$ $x=\frac{1}{t}$

 $$v=-c_1\frac{\sqrt{x^2+1}}{x}+c_2$$

$we$ $have$ $y=vx^2$

$so$ $putting$ $value$ $of$ $v$ $then$ $we$ $have$ 

$$y=-c_1x\sqrt{x^2+1}+c_2x^2$$

$$y=-c_1\sqrt{x^4+x^2}+c_2x^2$$

$Answer.$