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If y=a cos (logx)+b sin (logx)

prove that x²y∨2+xy∨1+y=0
can you explain y ^ 2 and y ^ 1 ?
what is V2 and V1 in the equation to prove?



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   y  = a cos (log x)  + b sin (log x)
x²y = ax² cos (logx)  + b x² sin (log x)
xy  =  ax cos (log x)  +  bx  sin (log x)
Differentiating y two time we get:
dy/dx  =  -a sin (log x) . 1/x  + b cos (log x)  . 1/x
           = 1/x [ -a sin (log x)  + b cos (log x)
d² y/ dx²  = a/x² sin (logx) - a/x² cos (log x)  - b/x² sin (log x)  - b/x² cos (log x)
               = a/x² [ sin (logx) - cos (logx)]  - b/x² [sin logx + cos logx]
x² d²y/dx²  =  a [ sin (logx) - cos (logx)]  - b [sin logx + cos logx]
x²y d²y/dx²  =  a² sin (logx) cos logx - a² cos² (logx)  - ab y cos logx + ab sin² logx
                             - ab cos logx sin logx - b²y sin logx
 xy dy/dx   =  -a² sin logx cos logx + ab cos² logx -ab sin² logx + b² sin logx cos logx
Add the terms in the above two and y, you get answer
0 0 0
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