Question

Solve:

$x\frac{dy}{dx}=y(logy-logx-1)$.

Answer 1

$x\frac{dy}{dx}=y(\ln y-\ln x-1)$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x}(\ln y-\ln x-1)$

$\Rightarrow \frac{1}{y}=\frac{dy}{dx}=\frac{\ln y}{x}-\left(\frac{1+\ln x}{x}\right)$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}-\frac{\ln y}{x}=-\left(\frac{1+\ln x}{x}\right)$.......(1)

Let $\ln y=t$

Differential w.r.t.x both sides

we get,

$\frac{1}{y}\frac{dy}{dx}=\frac{dt}{dx}$

So, eqn (1) becomes,

$\Rightarrow \frac{dt}{dx}-\frac{t}{x}=-\left(\frac{1+\ln x}{x}\right)$.......(2)

Above differential eqn is a linear differential equation of the form:-

$\frac{dy}{dx}+py=Q$

$\therefore I.F=e^{\int Pdx}=e^{\int -\frac{1}{x}dx}(\because P=\frac{-1}{x}ineqn(2\left)\right)Q=(-\frac{(1+\ln x)}{x})$

we know, $=e^{-\left(\ln x\right)}=e^{\ln x^{-1}}=e^{\ln \frac{1}{x}}=\frac{1}{x}(\because e^{\ln x}=x)$

$\therefore$ Solution is given by

$y\times I.F=\int (I.F.\times Q)dx+C$

So, solution for eqn (2) is given by

$t.\frac{1}{x}=\int \frac{-(1+\ln x)}{x}\times \frac{1}{x}dx+C$

$=\int \frac{-1}{x^2}dx-\int \frac{\ln x}{x^2}dx$

$t.\frac{1}{x}=\frac{1}{x}-I_1$.........(3)

Continued : $\to$

Let $I_1=\int \frac{\ln x}{x^2}dx$

Let $\ln x=u\Rightarrow x=e^4$

Diff we get $\frac{1}{x}dx=du$

$\therefore I_1=\int \frac{u}{(e^u{)}^2}du$

$=\int u.e^{-2u}du$

using ILATE Rule.

$I_1=u.\int e^{-2u}du-\int (\frac{du}{dv}.\int e^{-2u}du)du$

$=u\times \frac{e^{-2u}}{2}-\int 1\times \frac{e^{-2u}}{-2}du$

$=u\times \frac{e^{-2u}}{2}+\frac{1}{2}\int e^{-2u}du$

$=u\times \frac{e^{-2u}}{2}+\frac{1}{2}\times \frac{e^{-2u}}{-2}+c$

$=\frac{1}{2}[u{e}^{-2u}-\frac{1}{2}{e}^{-2u}]+c$

putting value of $u$, we get

$I_1=\frac{1}{2}[\ln x{e}^{-2\ln x}-\frac{1}{2}{e}^{\ln x}]+c$

$=\frac{1}{2}[\ln x\times e^{\ln x^{-2}}-\frac{1}{2}{e}^{\ln x^{-2}}]+c$

$=\frac{1}{2}[\ln x\times x^{-2}-\frac{1}{2}\times x^{-2}]+c$

$=\frac{1}{2}[\frac{\ln x}{x^2}-\frac{1}{2x^2}]+c$

$=\frac{1}{2}\left[\frac{2\ln x-1}{2}\right]+c$

$I_1=\frac{1}{4x^2}[\ln x^2-1]+c$

Now Putting value of $I_1$

in eqn (3), we get

$t.\frac{1}{x}=\frac{1}{x}-\frac{1}{4x^2}[\ln x^2-1]+c$

Putting value of $t$ we get.

$\frac{\ln y}{x}=\frac{1}{x}-\frac{1}{4x^2}[\ln x^2-\ln e]+c(\because \ln e=1)$

$\ln y=\frac{4x-\ln (x^2\times e)}{4x}+c$

final solution for $d$ eqn.