1.

If y = cos (lnx). Show that x2y2 + xy1 + y = 0.

Answer»

y = cos(lnx)

\(y_1 = \frac{dy}{dx} = \frac{-sin(lnx)}{x}\)

\(xy_1 = - sin(lnx)\)

\(y _2 = \frac{dy}{dx^2} = \frac{- cos(lnx)}{x^2}+ \frac{sin(lnx)}{x^2}\)

\(\therefore x^2y^2 = sin(lnx) - cos(lnx)\)

Now,

\(x^2y_2 + xy_1 + y = sin(lnx) - cos(lnx) - sin(lnx) + cos(lnx)\)

= 0

Hence Proved.



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