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`x(dy)/(dx)=y(logy-logx+1)` |
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Answer» Correct Answer - `y=xe^(cx), c gt 0` `x(dy//dx)=y(logy-logx+1)` or `(dy)/(dx)=y/x[logy/x+1]` Putting `y=vx`, we get `(dy)/(dx)=v+x(dv)/(dx)` and the given equation transforms to `v+x(dv)/(dx)=v[logv+1]` or `x(dv)/(dx)=vlogv` or `int(dv)/(dx)=vlogv` or `int(dv)/(vlogv)=int(dx)/x` or `loglogv=logx+logc, c gt 0` or `cx=log(y//x)` or `y=xe^(cx), c gt 0` |
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