y=a e^(-1/x)+b is a solution of (dy)/...

`y=a e^(-1/x)+b` is a solution of `(dy)/(dx)=y/(x^2),` then (a) `( b ) (c) a in R (d)` (e) (b) 0 (c) `( d ) (e) b=1( f )` (g) (d) `( h ) a (i)` (j) takes finite number of values

`x in R-{0}`

`b=0`

`b=1`

a takes finite number of values

Text Solution

Verified by Experts

The correct Answer is:

A, B

`(dy)/(dx)=y/x^(2)`
or `(dy)/(y) = (dx)/x^(2)`
or `"ln "y=-1/x+"ln "c` or `y/c=e^(-1/x)`
or `y=ce^(-1//x)`
Comparing with `y=ae^(-1//x)+b, a in R-{0},b=0`

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