If Differentiating with respect to x⇒ [ we can also write this as xy1 = -3sin(log x ) +4cos(log x )Differentiating with respect to x⇒ ⇒ ⇒ x2y2 + xy1 + y = 0Hence Proved
Differentiating with respect to x
⇒ [ we can also write this as xy1 = -3sin(log x ) +4cos(log x )
⇒
⇒ x2y2 + xy1 + y = 0
Hence Proved