Solve (1+y2)dx = (tan-1y - x)dy Share with your friends Share 21 Manbar Singh answered this The given differential equation is, 1 + y2 dx = tan-1y - x dy⇒dxdy = tan-1y - x1 + y2⇒dxdy + 11 + y2 . x = tan-1y1 + y2The above equation is linear and of the form of dxdy + Px = Q, whereP = 11 + y2 and Q =tan-1y1 + y2Integrating factor, IF = e∫P dy = e∫dy1+y2 = etan-1yNow, the required solution is given by,x × IF = ∫Q × IF dy + C⇒x etan-1y = ∫tan-1y1 + y2 × etan-1y dy + C ...........1Let I = ∫tan-1y1 + y2 × etan-1y dyput tan-1y = t11 + y2 dy = dtI = ∫t et dt=t × ∫etdt - ∫ddtt× ∫etdt dt=tet - ∫etdt=tet - et= tan-1y etan-1y - etan-1yNow, from 1, we getx etan-1y = tan-1y etan-1y - etan-1y + C⇒x = tan-1y - 1 + C e-tan-1y 91 View Full Answer
