xdx+ydyydx-xdy=xsin(x2+y2)y3 has the solution

x dx+ydysin(x2+y2)=(xy).(ydx−xdyy2)⇒ 12∫2xdx+2ydysin(x2+y2)=∫(xy).D(xy)Let x2+y2=u and xy=v⇒ 2xdx+2ydy=du, D(xy)=dv⇒ ∫1sin u du=2∫v dv⇒ log |Tan (x2+y22)|=(xy)2+c

xdx+ydyydx-xdy=xsin(x2+y2)y3 has the solution

x dx+ydysin(x2+y2)=(xy).(ydx−xdyy2)⇒ 12∫2xdx+2ydysin(x2+y2)=∫(xy).D(xy)Let x2+y2=u and xy=v⇒ 2xdx+2ydy=du, D(xy)=dv⇒ ∫1sin u du=2∫v dv⇒ log |Tan (x2+y22)|=(xy)2+c

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$\frac{xdx + ydy}{ydx - xdy} = \frac{xsin (x^{2} + y^{2})}{y^{3}}$ has the solution

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$\log |\tan (x^{2} + y^{2})| = \frac{y^{2}}{x^{2}} + c$

$\log |\tan (\frac{x^{2} + y^{2}}{2})| = \frac{x^{2}}{y^{2}} + c$

$\log |\tan (x^{2} + y^{2})| = \frac{x^{2}}{y^{2}} + c$

$\log |\tan (\frac{x^{2} + y^{2}}{2})| = \frac{y^{2}}{x^{2}} + c$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{x d x + y d y}{\sin (x^{2} + y^{2})} = (\frac{x}{y}) . (\frac{y d x - x d y}{y^{2}}) \\ \Rightarrow \frac{1}{2} \int \frac{2 x d x + 2 y d y}{\sin (x^{2} + y^{2})} = \int (\frac{x}{y}) . D (\frac{x}{y}) \\ L e t x^{2} + y^{2} = u a n d \frac{x}{y} = v \\ \Rightarrow 2 x d x + 2 y d y = d u, D (\frac{x}{y}) = d v \\ \Rightarrow \int \frac{1}{\sin u} d u = 2 \int v d v \\ \Rightarrow \log | T a n (\frac{x^{2} + y^{2}}{2}) | = {(\frac{x}{y})}^{2} + c \end{array}$