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Questions  

 Solution of the differential equation xdx+ydyxdxydy=y3x3 is given by (Base of all log=e ) 

a
32logyx+logx32+y32x32+tan−1yx32+c=0
b
23logyx+logx32+y32x32+tan−1yx+c=0
c
23logyx+logx+yx+tan−1y32x32+c=0
d
12logx3+y3+tan−1yx32+c=0

detailed solution

Correct option is D

We have xdx+ydyxdx−ydy=y3x3⇒dx32+dy32dx32−dy32=y32x32⇒du+dvdu−dv=vu Where u=x32 and v=y32⇒udu+udv=vdu−vdv⇒udu+vdv=vdu−udv⇒udu+vdvu2+v2=vdu−udvu2+v2⇒du2+v2u2+v2=−2dtan−1⁡vu On integrating we get log⁡u2+v2=−2tan−1⁡vu−2c⇒12log⁡x3+y3+tan−1⁡yx32+c=0

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