InterviewSolution
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InterviewSolution
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Find the general solution of the differential equations:`xlogx(dx)/(dy)+y=2/xlogx` |
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Answer» Given, `x log x(dy)/(dx)+y=(2)/(x)logx` `implies (dy)/(dx)+(y)/(xlogx)=(2)/(x^(2))` Comparing with differential equation `(dy)/(dx)+Py=Q` `P=(1)/(xlogx)` and `Q=(2)/(x^(2))` `I.F.=e^(intPdx)=e^(int(1dx)/(xlogx))` Let `logx=timplies(1)/(x)=(dt)/(dx)implies(dx)/(x)=dt` `:. I.F. =e^(int((1)/(t)dt)=e^(log|t|)=t=logx` `:.` Solution of given differential equation, `y*I.F.=intQ*I.F.dx+C` `impliesy*logx=int(2)/(x^(2))logxx+C` `implies ylogx=2int(logx*(1)/(x^(2)))dx+C` `=2[logxint(1)/(x^(2))dx-int{(d)/(dx)(logx)int(1)/(x^(2))dx}dx]+C` `=2[logx(-(1)/(x))-int{(1)/(x)*(-(1)/(x))}dx]+C` `=2(-(logx)/(x)+int(1)/(x^(2))dx)+C` `=2(-(logx)/(x)-(1)/(x))+C` `implies ylogx=-(2)/(x)(1+kogx)+C` |
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