Solve:
(π₯ β 2π πππ¦ + 3)ππ₯ + (2π₯ β 4π πππ¦ β 3)πππ π¦ππ¦ = 0
"\\begin{gathered}\n(x-2 \\sin y+3) d x+(2 x-4 \\sin y-3) \\cos y d y=0 \\\\\nu=\\sin y \\\\\nd u=\\cos y d y \\Rightarrow d y=d u \/ \\cos y\n\\end{gathered}"
Substituting in differential equation.
Β "\\begin{aligned}\n\n&(x-2 u+3) d x+(2 x-4 u-3) \\cos y \\frac{d u}{\\cos y}=0 \\\\\n\n& \\text { Integrating } \\\\\n\n\\Rightarrow & \\int(x-2 u+3) d x+\\int(2 x-4 u-3) d u=C \\\\\n\n\\Rightarrow & \\frac{x^{2}}{2}-2u x+3 x+\\frac{2 x^{2}}{2}-4u x-3 u=C \\\\\n\n\\Rightarrow & \\frac{3 x^{2}-64 x-3 u+3 x=C}{2}\\\\\\Rightarrow & \\frac{3 x^{2}}{3}-6 x \\sin y-3 \\sin y+3 x=C\n\n\\end{aligned}"
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