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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=(emtan1xy), 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+1vv(v+1).dv=dxx

dvvdvv+1=dxx
log vlog|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=emtan1xy
dydx=emtan1x(1+x2)y(1+x2)
dydx+y(1+x2)=emtan1x(1+x2)

P=1(1+x2),Q=emtan1x(1+x2)I.F.=ePdx=e1(1+x2)dx=etan1xThus the solution isyePdx=QePdxdxy×etan1x=emtan1x(1+x2).etan1x.dxy×etan1x=e(m+1)tan1x(1+x2).dx (i)e(m+)tan1x(1+x2)dx.......(ii)Let (m+1)tan1x=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)tan1x(m+1)Substituting in (i),y×etan1x=e(m+1)tan1x(m+1)+C........(iii)Putting y=1 and x=1, in the above equation,y×etan1x=e(m+1)tan1x(m+1)+C1×eπ4=e(m+1)π4(m+1)+CC=e(m+1)π4(m+1)eπ4

Particular solution of the D.E. is y×etan1x=e(m+1)tan1x(m+1)+e(m+1)π4(m+1)eπ4


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