Question

4(3x+y-2)dx-(3x+y)dy=0

Answer 1

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