InterviewSolution
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InterviewSolution
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`y dx+x log((y)/(x))dy-2x dy=0` |
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Answer» `y dx+x log((y)/(x))dy-2x dy=0` `implies y dx+[x log((y)/(x))-2x]dy=0` `implies (y)/(x)+[log((y)/(x))-2](dy)/(dx)=0` `implies (dy)/(dx)=[((y)/(x))/(2-log((y)/(x)))]" ....(1)"` माना y = vx (`because` समीकरण समघातीय है) `implies (dy)/(dx)=v+x(dv)/(dx)` समीकरण (1) में रखने पर, `v+x(dv)/(dx)=(v)/(2-logv)` `implies x(dv)/(dx)=(v)/(2-logv)-v` `=(vlogv-v)/(2-logv)` `implies (2-logv)/(v(logv-1))dv=(dx)/(x)` `impliesint(2-logv)/(v(logv-1))dv=int(dx)/(x)` `log v = t implies (1)/(v)dv=dt` `implies int(2-t)/(t-1)dt=int(dx)/(x)` `implies int(-1+(1)/(t-1))dt=int(dx)/(x)` `implies -t+log(t-1)=logx+logc` `implies-logv+log(logv-1)=log(cx)` `implies(logv-1)/(v)=cx` `implies logv-1=cxv` `implies log((y)/(x))-1=cy` |
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