Answer to Question #178632 in Differential Equations for LYRA

Question #178632

y(2x^2−xy+1)dx + (x−y)dy = 0


1
Expert's answer
2021-04-13T23:11:16-0400

dydx=y(2x2xy+1)yxSubstitute  y=vxxdvdx+v=vx(2x2vx2+1)vxvxdvdx=v(2x2vx2+1)v1vdvdx=v(2x2vx2+1)x(v1)vxdvdx=2vx2v2x2v2+2vx(v1)=(x2+1)(2vv2)x(v1)v1v(2v)  dv=x+1x  dxv1v(2v)  dv=x+1x  dxv12v  dv+v12(2v)  dv=x+1x  dxv12v  dvv12(v2)  dv=x+1x  dxv12v  dvv2+12(v2)  dv=x22+lnx+Cv2lnv2v2ln(v2)2=x22+lnx+Cln(v22v)2=x22+lnx+Cln(v22v)=x22lnx+Cv22v=Aex22lnx=Ax2ex2y2x22yx=Ax2ex2y22xy=Aex2y22xyAex2=0y=x±x2+Aex2is the solution to the ODE\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}\\


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