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Question

The particular solution of the differential equation xdydx=yln(yx),x0 is y=f(x). If f(1)=e2, then

A
f(x)=xe1+x2
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B
f(x)=xe1+x
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C
there are more than one tangent parallel to xaxis to the curve y=f(x)
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D
there is exactly one tangent parallel to xaxis to the curve y=f(x)
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Solution

The correct options are
B f(x)=xe1+x
D there is exactly one tangent parallel to xaxis to the curve y=f(x)
xdydx=yln(yx) (1)
dydx=yxln(yx)
Put y=vx
dydx=v+xdvdxv+xdvdx=vlnvxdvdx=v(lnv1)dvv(lnv1)=dxx
ln|ln(v)1|=ln|x|+ln|C|ln(v)1=Cxlnyx=1+Cx
Now, f(1)=e2C=1
The particular solution is y=xe1+x (2)

If tangent is parallel to xaxis, then
dydx=0,
e1+x+xe1+x=0x=1 (e1+x0)

Putting x=1 in equation (2),
y=1
So, only one tangent is parallel to xaxis to the curve y=f(x)

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