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Question

The solution of the differential equation xdx+ydyxdyydx=1x2y2x2+y2 is:

A
x2+y2=sin{tan1(yx)+c}
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B
x2+y2=cos{tan1(yx+c)}
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C
x2+y2=tan{sin1(yx+c)}
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D
y=xtan(c+sin1x2+y2)
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Solution

The correct options are
B x2+y2=sin{tan1(yx)+c}
D y=xtan(c+sin1x2+y2)
xdx+ydyxdyydx=1x2y2x2+y2
The above differential equation can be re-written as,
xdx+ydy1(x2+y2)=xdyydxx2+y2.................(1)
d(tan1(yx))=xdyydxx2+y2 and d(x2+y2)=2(xdx+ydy) .................(2)
multiplying both sides in the denominator of equation (1) with (x2+y2)
we get, xdx+ydyx2+y21(x2+y2)=xdyydxx2+y2 ...........................(3)
From (2) and (3), we have
12d(x2+y2)x2+y21(x2+y2)=d(tan1(yx))
Now, put x2+y2=t2 in the L.H.S. and we get,
12d(t2)t(1t2)=d(tan1(yx))
tdtt(1t2)=d(tan1(yx))
dt(1t2)=d(tan1(yx))
Now integrating both the sides i.e.
dt(1t2)=d(tan1(yx))
sin1t=tan1(yx)+C (where C is a constant)
Now, put the value of t, we get
C+sin1(x2+y2)=tan1(yx)........(4)
y=xtan(C+sin1(x2+y2))
Hence, option (D) is correct.
Again re-writing equation (4),
sin1(x2+y2)=tan1(yx)+C
(x2+y2)=sin(tan1(yx)+C)
Hence, option (A) is also correct.

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