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If x y-(log)e y=1 satisfies the equation...

If `x y-(log)_e y=1` satisfies the equation `x(y y_2+y1 2)-y_2+lambday y_1=0,t h e nlambda=` `-3` (b) 1 (c) 3 (d) none of these

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Given, equation is
`xy−logy=1`
On differentiating it w.r.t.x, we get
`xy′+y⋅1−1/y​⋅y′=0`
`⇒xyy′+y^2−y′=0`
Again, differentiating w.r.t. x, we get
`((xy−1)y′′+y′(xy′+y⋅1)+2yy′=0`
`⇒x(yy"+y(^′2))−y′′+3yy′=0`
...
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