Question

# Find the particular solution of the differential equation x2dy=(2xy+y2)dx, given that y= 1 when x = 1. OR Find the particular solution of the differential equation (1+x2)dydx=(emtan−1x−y), given that y =1 when x = 0.

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Solution

## x2dy=(2xy+y2)dx ⇒dydx=2xy+y2x2......(i) Let y=vx, dydx=v+xdvdx Substituting in (i), we get v+xdvdx=2vx2+v2x2x2 ⇒v+xdvdx=2v+v2 ⇒xdvdx=v+v2 ⇒dvv2+v=dxx Integrating both sides. ∫dvv2+v=∫dxx ⇒∫v+1−vv(v+1).dv=∫dxx ⇒∫dvv−∫dvv+1=∫dxx ⇒log v−log|v+1|=log x+log C ⇒log|vv+1|=log|Cx| ⇒log|yxyx+1|=log|Cx| ⇒yy+x=Cx [Removing logarithm in both sides] ∴ y=Cxy+Cx2,which is the general solution. Putting y=1 and x=1, 1=C+C ⇒2C=1 ⇒C=12 y=xy2+x22 ∴ 2y=xy+x2,which is the particular solution. OR (1+x2)dydx=emtan−1x−y ⇒dydx=emtan−1x(1+x2)−y(1+x2) ⇒dydx+y(1+x2)=emtan−1x(1+x2) P=1(1+x2),Q=emtan−1x(1+x2)I.F.=e∫Pdx=e∫1(1+x2)dx=etan−1xThus the solution isye∫Pdx=∫Qe∫Pdxdx⇒y×etan−1x=∫emtan−1x(1+x2).etan−1x.dx⇒y×etan−1x=∫e(m+1)tan−1x(1+x2).dx (i)∫e(m+)tan−1x(1+x2)dx.......(ii)Let (m+1)tan−1x=z(m+1)(1+x2)dx=dzdx(1+x2)=dz(m+1)Substituting in (ii),1(m+1)∫ezdz=ez(m+1)=e(m+1)tan−1x(m+1)Substituting in (i),⇒y×etan−1x=e(m+1)tan−1x(m+1)+C........(iii)Putting y=1 and x=1, in the above equation,⇒y×etan−1x=e(m+1)tan−1x(m+1)+C⇒1×eπ4=e(m+1)π4(m+1)+C∴C=e(m+1)π4(m+1)−eπ4 Particular solution of the D.E. is y×etan−1x=e(m+1)tan−1x(m+1)+e(m+1)π4(m+1)−eπ4

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