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`y=a e^(-1/x)+b`is asolution 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) takesfinite number of valuesA. `x in R-{0}`B. `b=0`C. `b=1`D. a takes finite number of values |
Answer» Correct Answer - 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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