($x^2 - y^2$)dx + 2xydy = 0 ⇒ $\frac{dy}{dx}$ = $\frac{y^2-x^2}{2xy}$ ... (1) It is a homogeneous differential equation. Let y = vx ...(2) ∴ $\frac{dy}{dx}$ = v + x $\frac{dv}{dx}$ ... (3) Substituting (2) and (3) in (1), we get: v + x $\frac{dv}{dx}$ = $\frac{v^2x^2-x^2}{2xvx}$ v + x $\frac{dv}{dx}$ = $\frac{x^2v^2-1}{2vx^2}$ - $\frac{v^2-1}{2v}$ $2v^2 + 2vx\frac{dv}{dx}$ = $v^2$ - 1 2vx $\frac{dv}{dx}$ = - $v^2$ - 1 $\left(\frac{2v}{v^2+1}\right)$ dv = - $\frac{dx}{x}$ Integrating both sides, we get: ∫ $\frac{2v}{v^2+1}$ dv = - ∫ $\left(\frac{1}{x}\right)$ dx log $\left|v^2+1\right|$ = - log |x| + log C log $\left|v^2+1\right|$ = log $\left|\frac{C}{x}\right|$ $v^2$ + 1 = $\frac{C}{x}$ $xv^2$ + 1 = C x $\left|\left(\frac{y}{x}\right)^2+1\right|$ = C $y^2+x^2$ = Cx ... 4 It is given that when x = 1, y = 1 $1^2+1^2$ = C(1) C = 2 Thus, the required solution is $y^2 + x^2$ = 2x. OR We need to solve the following differential equation $\frac{dy}{dx}$ = $\frac{x(2y-x)}{x(2y+x)}$ $\frac{dy}{dx}$ = $\frac{2y-x}{2y+x}$ ... (1) It is a homogeneous differential equation. Let y = vx ...(2) ∴ $\frac{dy}{dx}$ = v + x $\frac{dv}{dx}$ ... 3 Substituting (2) and (3) in (1), we get: v + x $\frac{dv}{dx}$ = $\frac{x(2v-1)}{x(2v+1)}$ x $\frac{dv}{dx}$ = $\frac{v-1}{2v+1}$ - v x $\frac{dv}{dx}$ = $\frac{-2v^2+v-1}{2v+1}$ $\left(\frac{2v+1}{-2v^2+v-1}\right)$ dv = $\left(\frac{1}{x}\right)$ dx $\left(\frac{2v+1}{2v^2-v+1}\right)$ dv = $\left(-\frac{1}{x}\right)$ dx Integrating both sides, $\int \frac{1}{2} \left(\frac{4v-1+3}{2v^2-v+1}\right)$ dv = ∫ $\left(-\frac{1}{x}\right)$ dx ∫ $\frac{1}{2} \left(\frac{4v-1}{2v^2-v+1}\right)$ dv + ∫ $\frac{3}{2} \left(\frac{1}{2v^2-v+1}\right)$ dv = ∫ $\left(-\frac{1}{x}\right)$ dx ∫ $\frac{1}{2}\left(\frac{4v-1}{2v^2-v+1}\right)$ dv + ∫ $\frac{3}{4} \left|\frac{1}{v^2 - \frac{v}{2} + \frac{1}{2}}\right|$ dv = f $\left(-\frac{1}{x}\right)$ dx 1/2 log $\left|2v^2 - v + 1\right|$ + $\frac{3}{4} \int \left(\frac{1}{v^2 - \frac{v}{2} + \frac{1}{16} + \frac{7}{16}}\right)$ dv = - log |x| + C 1/2 log $\left|2v^2 - v + 1\right|$ + $\frac{3}{4}$ ∫ $\frac{dv}{\left(v - \frac{1}{4}\right)^2 + \left(\frac{\sqrt{7}}{4}\right)^2}$ = - log |x| + C 1/2 log $\left|2v^2 - v + 1\right|$ + $\frac{3}{4} \times \frac{4}{\sqrt{7}} \tan^{-1}\left(\frac{v - \frac{1}{4}}{\frac{\sqrt{7}}{4}}\right)$ = - log |x| + C 1/2 log $\left|2v^2 - v + 1\right|$ + $\frac{3}{\sqrt{7}} \tan^{-1}\left(\frac{4v-1}{\sqrt{7}}\right)$ = C - log |x| Put v = $\frac{y}{x}$ $\frac{1}{2}$ log $\left|2\left(\frac{y}{x}\right)^2 - \frac{y}{x} + 1\right|$ + $\frac{3}{\sqrt{7}} \tan^{-1}\left(\frac{\frac{4y}{x} - 1}{\sqrt{7}}\right)$ = C - log |x| $\frac{1}{2}$ log $\left|\frac{2y^2 - xy + x^2}{x^2}\right|$ + $\frac{3}{\sqrt{7}} \tan^{-1}\left(\frac{4y-x}{\sqrt{7}x}\right)$ = C - log |x| ... 4 Now y = 1 when x = 1 $\frac{1}{2}$ log $\left|\frac{2(1)^2 - 1(1) + 1^2}{1^2}\right|$ + $\frac{3}{\sqrt{7}} \tan^{-1}\left|\frac{4(1)-1}{\sqrt{7}(1)}\right|$ = C - log |1| $\frac{1}{2}$ log 2 + $\frac{3}{\sqrt{7}} \tan^{-1} \left(\frac{3}{\sqrt{7}}\right)$ = C ... (5) Therefore, form (4) and (5) we get: $\frac{1}{2}$ log $\left|\frac{2y^2 - xy + x^2}{x^2}\right|$ + $\frac{3}{\sqrt{7}} \tan^{-1}\left(\frac{4y-x}{\sqrt{7}x}\right)$ = $\frac{1}{2}$ log 2 + $\frac{3}{\sqrt{7}} \tan^{-1} \left(\frac{3}{\sqrt{7}}\right)$ - log |x| $\frac{1}{2}$ log $\left|\frac{2y^2 - xy + x^2}{x^2}\right|$ - $\frac{1}{2}$ log 2 + log |x| = $\frac{3}{\sqrt{7}} \left[ \tan^{-1}\frac{3}{\sqrt{7}} - \tan^{-1}\left(\frac{4y-x}{\sqrt{7}x}\right) \right]$ $\frac{1}{2}$ log $\left|\frac{2y^2 - xy + x^2}{2x^2} \cdot x^2\right|$ = $\frac{3}{\sqrt{7}} \tan^{-1}\left(\frac{\frac{3x-4y+x}{\sqrt{7}x}}{1 + \frac{3(4y-x)}{7x}}\right)$ $\frac{1}{2}$ log $\left|\frac{2y^2 - xy + x^2}{2}\right|$ = $\frac{3}{\sqrt{7}} \tan^{-1}\left(\frac{\frac{4(x-y)}{\sqrt{7}x}}{\frac{7x+12y-3x}{7x}}\right)$ $\frac{1}{2}$ log $\left|\frac{2y^2 - xy + x^2}{2}\right|$ = $\frac{3}{\sqrt{7}} \tan^{-1}\left(\frac{\sqrt{7}(x-y)}{x+3y}\right)$