(y2−1)dx−2dy=0y2−1−2y′=0y2−1=2y′2y2−1 dy=dx∫2y2−1 dy=∫dx12⋅2ln(y−1y+1)=x+Cy−1y+1=ex+Cy−1y+1=Aexy−1=A(y+1)exy(1−Aex)=Aex+1y=1+Aex1−Aex\displaystyle (y² -1)\mathrm{d}x -2\mathrm{d}y =0\\ y^2 - 1 - 2y' = 0\\ y^2 - 1 = 2y' \\ \frac{2}{y^2 - 1} \,\mathrm{d}y = \mathrm{d}x\\ \int\frac{2}{y^2 - 1} \,\mathrm{d}y = \int\mathrm{d}x\\ \frac{1}{2}\cdot 2 \ln\left(\frac{y - 1}{y + 1}\right) = x + C\\ \frac{y - 1}{y + 1} = e^{x + C}\\ \frac{y - 1}{y + 1} = Ae^{x}\\ y - 1 = A(y + 1)e^{x}\\ y(1 - Ae^{x}) = Ae^{x} + 1 \\ y = \frac{1 + Ae^{x}}{1 - Ae^{x}}(y2−1)dx−2dy=0y2−1−2y′=0y2−1=2y′y2−12dy=dx∫y2−12dy=∫dx21⋅2ln(y+1y−1)=x+Cy+1y−1=ex+Cy+1y−1=Aexy−1=A(y+1)exy(1−Aex)=Aex+1y=1−Aex1+Aex
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments