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Question
Find the particular solution of the differential equation x2dy = (2xy + y2) dx, given that y = 1 when x = 1.
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Solution
x2dy = (2xy + y2)dx
`=>dy/dx=(2xy+y^2)/x^2.......(i)`
Let y=vx,
`dy/dx=v+xdv/dx`
Substituting in (i), we get
`v+x (dv)/dx=(2vx^2+v^2x^2)/x^2`
`=>v+x (dv)/dx=2v+v^2`
`=>x (dv)/dx=v^2+v`
`=>(dv)/(v^2+v)=dx/x`
integrating both sides
`=>int(dv)/(v^2+v)=intdx/x`
`=>(v+1-v)/(v(v+1))dv=intdx/x`
`=>logv-log|v+1|=logx+logC`
`=>log|v/(v+1)|=log|Cx|`
`=>log|(y/x)/(y/x+1)|=log|Cx|`
`=>y/(y+x)=Cx` [Removing logarithm in both sides]
`therefore y=Cxy+Cx^2` ,which is the general solution.
Putting y=1 and x=1,
`1=C + C`
`=>2C=1`
`=>c=1/2y`
`=(xy)/2+x^2/2`
`therefore 2y=xy+x^2,` which is the particular solution.
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