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Solve `x(dy)/(dx)=y(logy-logx+1)` |
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Answer» Given, `" "x(dy)/(dx)=y(logy-logx+1)` `rArr" "x(dy)/(dx)=ylog((y)/(x)+1)` `rArr" "(dy)/(dx)=(y)/(x)(log""(y)/(x)+1)" "`...(i) which is a homogeneous equation. Put `" "(y)/(x)=v or y=vx` `therefore" "(dy)/(dx)=v+x(dv)/(dx)` On substituting these values in Eq. (i), we get `rArr" "x(dv)/(dx)=v(logv+1-1)` `rArr" "x(dv)/(dx)=v(logv)` `rArr" "(dv)/(vlogv)=(dx)/(x)` On integrating both sides, we get `" "int(dv)/(vlogv)=int(dx)/(x)` On putting logv=u in LHS integral, we get `" "(1)/(v)*dv=du` `" "int(du)/(u)=int(dx)/(x)` `rArr" "logu=logx+logC` `rArr" "logu=logCx` `rArr" "u=Cx` `rArr" "logv=Cx` `rArr" "log((y)/(x))=Cx` |
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