Solve xdx + ydy = frac{xdy - ydx}{x^2 + y^2} | vedclass

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(A) The given differential equation is $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$.We know that $d(x^2 + y^2) = 2xdx + 2ydy$,which implies $xdx + ydy =

Vedclass · Accepted Answer

$\frac{1}{2}(x^2 + y^2) = 	an^{-1}(y/x) + c$

Vedclass · Answer

(A) The given differential equation is $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$.We know that $d(x^2 + y^2) = 2xdx + 2ydy$,which implies $xdx + ydy = \frac{1}{2} d(x^2 + y^2)$.Also,we know that $d(	an^{-1}(y/x)) = \frac{1}{1 + (y/x)^2} \cdot d(y/x) = \frac{x^2}{x^2 + y^2} \cdot \frac{xdy - ydx}{x^2} = \frac{xdy - ydx}{x^2 + y^2}$.Substituting these into the given equation,we get:$\frac{1}{2} d(x^2 + y^2) = d(	an^{-1}(y/x))$.Integrating both sides,we get:$\frac{1}{2}(x^2 + y^2) = 	an^{-1}(y/x) + c$.

Solve $xdx + ydy = \frac{xdy - ydx}{x^2 + y^2}$

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