"\\displaystyle\n\n4(3x+y-2)\\mathrm{d}x -(3x+y)\\mathrm{d}y=0\\\\\n\n4(3x+y-2) -(3x+y)\\frac{\\mathrm{d}y}{\\mathrm{d}x} =0\\\\\n\n\\textsf{Substitute}\\, v = 3x + y\\\\\n\n\n\\frac{\\mathrm{d}v}{\\mathrm{d}x} = 3 + \\frac{\\mathrm{d}y}{\\mathrm{d}x}\\\\\n\n\n\n\\frac{\\mathrm{d}v}{\\mathrm{d}x} - 3 = \\frac{\\mathrm{d}y}{\\mathrm{d}x}\\\\\n\n\n\n\\implies 4(v -2) - v\\left(\\frac{\\mathrm{d}v}{\\mathrm{d}x} - 3\\right) =0\\\\\n\n\n\n7v - 8 - v\\frac{\\mathrm{d}v}{\\mathrm{d}x}=0\\\\\n\n\n7v - 8 = v\\frac{\\mathrm{d}v}{\\mathrm{d}x}\\\\\n\n\n\\frac{7v - 8}{v} = \\frac{\\mathrm{d}v}{\\mathrm{d}x}\\\\\n\n \\frac{\\mathrm{d}v}{\\mathrm{d}x} = \\frac{7v - 8}{v}\\\\\n\n\n\\frac{v\\,\\mathrm{d}v}{(7v - 8)\\mathrm{d}x} = 1\\\\\n\n\n\\textsf{Integrating both sides}\\, wrt.\\, x\\\\\n\n\\int \\frac{v\\,\\mathrm{d}v}{(7v - 8)\\mathrm{d}x} \\cdot \\mathrm{d}x = \\int\\, \\mathrm{d}x\\\\\n\n\n\\int \\frac{v\\mathrm{d}v}{7v - 8}\\, \\mathrm{d}x = x\\\\\n\n\nx = \\frac{1}{7}\\int \\frac{7v - 8 + 8}{7v - 8}\\cdot \\mathrm{d}v\\\\\n\n\nx = \\int \\left(\\frac{1}{7} + \\frac{8}{7(v - 8)}\\right)\\mathrm{d}v\\\\\n\n\n\nx = \\frac{v}{7} + \\frac{8}{7}\\ln(v - 8 ) + C\\\\\n\n\n\n7x = (3x + y) + 8\\ln(3x + y - 8 ) + C\\\\\n\n\n\n(4x - y) - C = 8\\ln(3x + y - 8 )\\\\\n\n\n\ne^{\\frac{(4x - y) - C}{8}} = 3x + y - 8\\\\\n\n\n\ny = 8 - 3x + e^{\\frac{(4x - y) - C}{8}}\\\\\n\n\ny = 8 - 3x + Ae^{\\frac{(4x - y)}{8}}, \\hspace{0.5cm}\\{\\textsf{where}\\, A = e^{\\frac{-C}{8}}\\}\\\\\n\n\\therefore y = 8 - 3x + Ae^{\\frac{(4x - y)}{8}} \\, \\, \\textsf{is the solution the ODE}."
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