"\\displaystyle\ny\\mathrm{d}x+ (x-y+x-2y)\\mathrm{d}y =0 \\\\\n\ny\\mathrm{d}x+ (2x - 3y)\\mathrm{d}y =0 \\\\\n\n\ny\\mathrm{d}x+ (2x - 3y)\\mathrm{d}y =0 \\\\\n\n\\frac{\\mathrm{d}x}{\\mathrm{d}y} = \\frac{3y - 2x}{y}\\\\\n\n\n\\frac{\\mathrm{d}x}{\\mathrm{d}y} = 3 - \\frac{2x}{y}\\\\\n\n\n\n\\frac{\\mathrm{d}x}{\\mathrm{d}y} + \\frac{2x}{y} = 3\\, \\textsf{as a linear form}\\\\\n\n\n\\textsf{Integrating factor}\\,\\, (IF) = e^{\\int \\frac{2}{y}\\,\\mathrm{d}y} = e^{2\\ln{y}} = e^{\\ln(y^2)} = y^2\\\\\n\n\n\\frac{\\mathrm{d}x}{\\mathrm{d}y} + \\frac{2x}{y} = 3\\\\\n\n\nxIF = \\int 3 IF\\, \\mathrm{d}y\\\\\n\nxy^2 = \\int 3y^2 \\, \\mathrm{d}y\\\\\n\n\nxy^3 = y^3 + C\\\\\n\nx = 1 + Cy^{-3}\\, \\textsf{is the solution to the ODE}"
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