If x(t) is a solution of (t + 1)dx=(2x+(t + 1)4 ) dt, x(0)=2  then limt→1 x(t) is

dxdt−2t+1x=(t+1)3 The I. F. is e−2∫dtt+1=1(t+1)2The equation reduces to  ddtx1(t+1)2=(t+1)⇒x(t+1)2=(t+1)22+C. Since x(0)=h2so , C=32. Thus x(t+1)2=(t+1)22+32 ⇒x=(t+1)42+32(t+1)2.  limt→1 x(t)=8+6=14