MCQ
The solution of the differential equation ${x^2}dy = - 2xydx$ is
- A$x{y^2} = c$
- B${x^2}{y^2} = c$
- ✓${x^2}y = c$
- D$xy = c$
On integrating, $\log y = - 2\log x + \log c$
==> $\log y = \log {x^{ - 2}} + \log c$ ==> $\log y{x^2} = \log c$ or $y{x^2} = c$.
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$x\,\, + \,\,y\,\, = \,\,\frac{{2\pi }}{3},\,{\rm{cos}}\,{\rm{x + }}\,{\rm{ cos}}\,{\rm{y}}\,{\rm{ = }}\,\frac{3}{2},$ where $x$ and $y$ are real in