The solution of the differential equation x−y arc $$tanyxdx+x$$ $$arcTanyxdy=0$$ is of the form $$logx4x2+y2+k$$ $$tan$$−$$1yx$$ $$=$$ c then the value of k is

Question

The solution of the differential equation x−y  arc   $$tanyxdx+x$$  $$arcTanyxdy=0$$ is of the form $$logx4x2+y2+k$$ $$tan$$−$$1yx$$   $$=$$ c then the value of k is

Infinity Learn · Accepted Answer

The given equation can rewritten as $$dydx=ytan$$−$$1$$⁡yx−xxtan−$$1$$⁡$$yx=yx$$−$$1tan$$−$$1$$⁡$$y/xv+xdvdx=v$$−$$1tan$$−$$1$$⁡v $$($$ on substituting $$y=vx)$$⇒∫$$tan$$−$$1$$⁡$$vdv=$$−∫dxxvtan−$$1$$⁡v−∫$$v11+v2dv+ln$$⁡$$x=c$$   by parts⇒ln⁡x−$$12$$∫$$2v1+v2dv+yxtan$$−$$1$$⁡$$yx=c$$⇒$$2ln$$⁡x−ln⁡$$1+v2+2yxtan$$−$$1$$⁡$$yx=c$$⇒$$2ln$$⁡x−ln⁡$$x2+y2x2+2yxtan$$−$$1$$⁡$$yx=c$$⇒ln⁡$$x4x2+y2+2yxtan$$−$$1$$⁡$$yx=c$$

The solution of the differential equation $(x - y a r c \tan \frac{y}{x}) d x + (x a r c T a n \frac{y}{x}) d y = 0$ is of the form $\log |\frac{x^{4}}{x^{2} + y^{2}}| + k \tan^{- 1} \frac{y}{x}$ = c then the value of k is

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The solution of the differential equation x−y arc tanyxdx+x arcTanyxdy=0 is of the form logx4x2+y2+k tan−1yx = c then the value of k is

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The solution of the differential equation $(x - y a r c \tan \frac{y}{x}) d x + (x a r c T a n \frac{y}{x}) d y = 0$ is of the form $\log |\frac{x^{4}}{x^{2} + y^{2}}| + k \tan^{- 1} \frac{y}{x}$ = c then the value of k is