The solution of the differential equation xdx+ydy+(xdy-ydx/x2+y2)=0, is

Question

The solution of the differential equation xdx+ydy+(xdy-ydx/x2+y2)=0, is (A) y=x tan((x2+y2+C/2)) (B) x=y tan((x2+y2+C/2)) (C) y=x tan((C-x2-y2/2)) (D) none of these

Tardigrade Admin · Accepted Answer

Given, xdx+ydy+(xdy-ydx/x2+y2)=0 ⇒ (1/2)d(x2+y2)+d(tan-1 (y/x))=0 ⇒ (1/2)(x2+y2)+tan-1 (y/x)=(C/2) (Integrating) ⇒ (C-x2-y2/2)=tan-1 (y/x) ⇒ y=x tan((C-x2-y2/2))

Q. The solution of the differential equation
$x d x + y d y + \frac{x d y - y d x}{x ^{2} + y ^{2}} = 0$ , is

$y = x t an (\frac{x ^{2} + y ^{2} + C}{2})$

$x = y t an (\frac{x ^{2} + y ^{2} + C}{2})$

$y = x t an (\frac{C - x ^{2} - y ^{2}}{2})$

none of these

Q. The solution of the differential equation xdx+ydy+x2+y2xdy−ydx​=0, is

y=xtan(2x2+y2+C​)

x=ytan(2x2+y2+C​)

y=xtan(2C−x2−y2​)

none of these

Given, xdx+ydy+x2+y2xdy−ydx​=0 ⇒21​d(x2+y2)+d(tan−1xy​)=0 ⇒21​(x2+y2)+tan−1xy​=2C​ (Integrating) ⇒2C−x2−y2​=tan−1xy​ ⇒y=xtan(2C−x2−y2​)

Q. The solution of the differential equation
$x d x + y d y + \frac{x d y - y d x}{x ^{2} + y ^{2}} = 0$ , is

$y = x t an (\frac{x ^{2} + y ^{2} + C}{2})$

$x = y t an (\frac{x ^{2} + y ^{2} + C}{2})$

$y = x t an (\frac{C - x ^{2} - y ^{2}}{2})$