can you explain y ^ 2 and y ^ 1 ?

what is V2 and V1 in the equation to prove?

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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

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