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Q.

The solution of y dx  x dy + logx dx =0 is

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a

y - logx  1 = cx

b

x + logy + 1= cx

c

y + logx + 1 = cx

d

y + logx  1 = cx

answer is C.

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Detailed Solution

y dx  x dy + logx dx =0
   xdy-ydxx2=logx dxx2     log xx2=ddx yx

 log x x2 dx= dyx log xx2 dx=yx+C1 Let log x=t  1x dx=dt x=et 1x=e-t  t e-t dt=-t e-t-e-t =-1+tet=--(1+log x)x  -1+log xx=yx+C1  1+log x+y+C1x=0 Put C1=-C  1+log x+y=Cx

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The solution of y dx – x dy + logx dx =0 is