Question

y(2x^2−xy+1)dx + (x−y)dy = 0

Answer 1

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y(2x^2 - xy + 1)}{y - x} \\ \textsf{Substitute}\,\, y = vx\\ x\frac{\mathrm{d}v}{\mathrm{d}x} + v = \frac{vx(2x^2 - vx^2 + 1)}{vx - v}\\ x\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{v(2x^2 - vx^2 + 1)}{v - 1} - v\\ \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{v(2x^2 - vx^2 + 1)}{x(v - 1)} - \frac{v}{x}\\ \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{2vx^2 - v^2x^2 - v^2 + 2v}{x(v - 1)} = \frac{(x^2 + 1)(2v - v^2)}{x(v - 1)}\\ \frac{v - 1}{v(2 - v)} \,\,\mathrm{d}v = x + \frac{1}{x} \,\, \mathrm{d}x\\ \int\frac{v - 1}{v(2 - v)} \,\,\mathrm{d}v = \int x + \frac{1}{x} \,\, \mathrm{d}x\\ \int\frac{v - 1}{2v} \,\,\mathrm{d}v + \int\frac{v - 1}{2(2 - v)} \,\,\mathrm{d}v = \int x + \frac{1}{x} \,\, \mathrm{d}x\\ \int\frac{v - 1}{2v} \,\,\mathrm{d}v - \int\frac{v - 1}{2(v - 2)} \,\,\mathrm{d}v = \int x + \frac{1}{x} \,\, \mathrm{d}x\\ \int\frac{v - 1}{2v} \,\,\mathrm{d}v - \int\frac{v - 2 + 1}{2(v - 2)} \,\,\mathrm{d}v = \frac{x^2}{2} + \ln{x} + C\\ \frac{v}{2} - \frac{\ln{v}}{2} - \frac{v}{2} - \frac{\ln{(v - 2)}}{2}= \frac{x^2}{2} +\ln{x} + C\\ -\frac{\ln{\left(v^2 - 2v\right)}}{2}= \frac{x^2}{2} + \ln{x} + C\\ \ln\left(v^2 - 2v\right) = -x^2 - 2\ln{x} + C \\ v^2 - 2v = Ae^{-x^2 - 2\ln{x}} = \frac{A}{x^2} e^{-x^2}\\ \frac{y^2}{x^2} - \frac{2y}{x} =\frac{A}{x^2} e^{-x^2} \\ y^2 - 2xy = A e^{-x^2}\\ y^2 - 2xy - Ae^{-x^2} = 0\\ \therefore y = x \pm \sqrt{x^2 + Ae^{-x^2}} \textsf{is the solution to the ODE}\\$