1.The general solution of the differential equation (tan−1y−x) dy=(1+y2) dx(\tan^{-1} y - x)\,dy = (1 + y^2)\,dx(tan−1y−x)dy=(1+y2)dx isa.x=(tan−1y+1)+Ce−tan−1yx = (\tan^{-1} y + 1) + Ce^{-\tan^{-1} y}x=(tan−1y+1)+Ce−tan−1yb.x=(tan−1y−1)+Ce−tan−1yx = (\tan^{-1} y - 1) + Ce^{-\tan^{-1} y}x=(tan−1y−1)+Ce−tan−1yc.x=(tan−1x−1)+Ce−tan−1xx = (\tan^{-1} x - 1) + Ce^{-\tan^{-1} x}x=(tan−1x−1)+Ce−tan−1xd.x=(tan−1x+1)+Ce−tan−1xx = (\tan^{-1} x + 1) + Ce^{-\tan^{-1} x}x=(tan−1x+1)+Ce−tan−1xLogin to continueOnly logged in users canattempt or see the solution.
