Question

यदि y = 3cos(log x) + 4 sin(log x) है तो दर्शाइए कि x^{2}y_{2} + xy_{2} + y = 0

Answer 1

x²y₂ + xy₁  + y  = 0 , यदि y = 3Cos(logx) + 4Sin(logx)

Step-by-step explanation:

प्रश्न में सुधार , दर्शाइए कि x²y₁  +  xy₂ + y = 0

y = 3Cos(logx) + 4Sin(logx)

dy/dx = 3(-Sin(logx))(1/x)  + 4Cos(logx)(1/x)

dy/dx = (4Cos(logx) - 3Sin(logx))/x

y₁ = dy/dx

=> xy₁ = 4Cos(logx) - 3Sin(logx)

d²y/dx² = (4(-Sin(logx))(1/x) - 3Cos(logx)(1/x))/x  + (4Cos(logx) - 3Sin(logx))(-1/x²)

y₂ = d²y/dx²

=> x²y₂ = -7Cos(logx) -Sin(logx)

x²y₂ + xy₁  + y  = 0

LHS

=  x²y₂ + xy₁  + y

=  -7Cos(logx) -Sin(logx)  + 4Cos(logx) - 3Sin(logx) +  3Cos(logx) + 4Sin(logx)

= 0

= RHS

=>  x²y₂ + xy₁  + y  = 0

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