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The solution of differential equation `(1+x) y dx +(1-y) x dy = 0 ` isA. ` log_(e)(xy)+x-y=C`B. ` log _(e)(x/y)+x+y = C`C. ` log_(e) (x/y)-x+y = C`D. `log_(e) (xy)-x+y = C` |
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Answer» Correct Answer - a Given , `(1+x) y dx +(1-y) x dy = 0 ` ` rArr ((1-y))/y dy +((1+x))/x dx = 0 ` On integrating both sides , we get ` rArr int (1/y -1) dy +int (1/x +1) dx = 0 ` ` rArr log_(e)y - y + log_(e) x + x =C` ` rArr log_(e) (xy) +x-y = C` |
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