If x=1−y1+y then x+1d2ydx2+3y+1ydydx=

x=1−y1+y⇒1=1+y−12ydydx−1−y12ydydx1+y2⇒1=1+y−12y−1−y12y1+y2dydxdydx=1+y212y−1−y−1+y       =−y1+y2Again differentiate both sides ⇒d2ydx2=−y21+y12y−1+y212ydydx=12y−2y−2y−1−y−2ydydx⇒d2ydx2+3y+4y+12ydydx=0⇒d2ydx2+3y+1y+12ydydx=0⇒2y+1d2ydx2+3y+1ydydx=0⇒x+1d2ydx2+3y+1ydydx=0    ∵x+1=1−y1+y+1=2y+1