Question

# Solve:xdydx=y(logy−logx−1).

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Solution

## xdydx=y(lny−lnx−1)⇒1ydydx=1x(lny−lnx−1)⇒1y=dydx=lnyx−(1+lnxx)⇒1ydydx−lnyx=−(1+lnxx).......(1)Let lny=tDifferential w.r.t.x both sideswe get,1ydydx=dtdxSo, eqn (1) becomes,⇒dtdx−tx=−(1+lnxx).......(2)Above differential eqn is a linear differential equation of the form:-dydx+py=Q∴I.F=e∫Pdx=e∫−1xdx(∵P=−1x in eqn(2))Q=(−(1+lnx)x)we know, =e−(lnx)=elnx−1=eln1x=1x(∵elnx=x)∴ Solution is given byy×I.F=∫(I.F.×Q)dx+CSo, solution for eqn (2) is given byt.1x=∫−(1+lnx)x×1xdx+C=∫−1x2dx−∫lnxx2dxt.1x=1x−I1.........(3)Continued : →Let I1=∫lnxx2dxLet lnx=u⇒x=e4Diff we get 1xdx=du∴I1=∫u(eu)2du=∫u.e−2uduusing ILATE Rule.I1=u.∫e−2udu−∫(dudv.∫e−2udu)du=u×e−2u2−∫1×e−2u−2du=u×e−2u2+12∫e−2udu=u×e−2u2+12×e−2u−2+c=12[ue−2u−12e−2u]+cputting value of u, we getI1=12[lnxe−2lnx−12elnx]+c=12[lnx×elnx−2−12elnx−2]+c=12[lnx×x−2−12×x−2]+c=12[lnxx2−12x2]+c=12[2lnx−12]+cI1=14x2[lnx2−1]+cNow Putting value of I1in eqn (3), we gett.1x=1x−14x2[lnx2−1]+cPutting value of t we get.lnyx=1x−14x2[lnx2−lne]+c(∵lne=1)lny=4x−ln(x2×e)4x+cfinal solution for d eqn.

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