Question

Solution of $\frac{y+\sin x{\cos }^2\left(xy\right)}{{\cos }^2\left(xy\right)}dx+(\frac{x}{{\cos }^2\left(xy\right)}+\sin y)dy=0$

• A. $\tan \left(xy\right)+\cos x+\cos y=c$
• B. $\tan \left(xy\right)-\cos x-\cos y=c$
• C. $\tan \left(xy\right)+\cos x-\cos y=c$
• D. $\tan \left(xy\right)-\cos x+\cos y=c$

Answer 1

Answer: (B)

$\tan \left(xy\right)-\cos x-\cos y=c$

$\frac{y+\sin x{\cos }^2\left(xy\right)}{{\cos }^2\left(xy\right)}dx+(\frac{x}{{\cos }^2\left(xy\right)}+\sin y)dy=0$

multiplying by ${\cos }^2\left(xy\right)$ in $LHS$ & $RHS$

$[y+\sin x{\cos }^2(xy\left)\right]dx+xdy+\sin y{\cos }^{2}xydy=0$

$\Rightarrow ydx+\sin x{\cos }^2\left(xy\right)dx+dy+\sin y{\cos }^{2}xydy=0$

$\Rightarrow (xdy+ydx)+{\cos }^{2}xy[\sin xdx+\sin ydy]=0$

$\Rightarrow d\left(xy\right)+{\cos }^{2}xy[\sin xdx+\sin ydy]=0$

diving $LHS$ & $RHS$ by ${\cos }^{2}xy$

$\Rightarrow \int \frac{d\left(xy\right)}{{\cos }^2\left(xy\right)}+\int \sin dx+\int \sin ydy=-$

$\Rightarrow \tan \left(xy\right)-\cos x-\cos y-C=0$

$\Rightarrow \tan \left(xy\right)-\cos x-\cos y=C....\left(B\right)$