The given differential equation is,
a m p ; ( y − 2 log x ) d x + ( x log x 2 ) d y = 0 a m p ; ⇒ d y d x = ( 2 log x − y ) 2 x log x a m p ; ⇒ d y d x + y 2 x log x = 1 x \begin{aligned}& (y-2 \log x) d x+\left(x \log x^2\right) d y=0 \\\\& \Rightarrow \frac{d y}{d x}=\frac{(2 \log x-y)}{2 x \log x} \\\\& \Rightarrow \frac{d y}{d x}+\frac{y}{2 x \log x}=\frac{1}{x}\end{aligned} am p ; ( y − 2 log x ) d x + ( x log x 2 ) d y = 0 am p ; ⇒ d x d y = 2 x log x ( 2 log x − y ) am p ; ⇒ d x d y + 2 x log x y = x 1 It is a linear differential equation.
∴ I.F. = e ∫ 1 2 x log x d x \therefore \text { I.F. }=e^{\int \frac{1}{2 x \log x} d x} ∴ I.F. = e ∫ 2 x l o g x 1 d x Put
log x = t ⇒ 1 x d x = d t \log x=t \Rightarrow \frac{1}{x} d x=d t log x = t ⇒ x 1 d x = d t ∴ I.F. = e ∫ 1 2 t d t = e log ( t ) 1 2 = t = log x \therefore \text { I.F. }=e^{\int \frac{1}{2 t} d t}=e^{\log (t)^{\frac{1}{2}}}=\sqrt{t}=\sqrt{\log x} ∴ I.F. = e ∫ 2 t 1 d t = e l o g ( t ) 2 1 = t = log x So, required solution is,
a m p ; y log x = ∫ log x x d x a m p ; log x = v ⇒ 1 x d x = d v a m p ; ⇒ y log x = ∫ v d v + C a m p ; ⇒ y log x = 2 v 3 / 2 3 + C a m p ; ⇒ y log x = 2 3 ( log x ) 3 / 2 + C \begin{aligned}& y \sqrt{\log x}=\int \frac{\sqrt{\log x}}{x} d x \\\\& \log x=v \Rightarrow \frac{1}{x} d x=d v \\\\& \Rightarrow y \sqrt{\log x}=\int \sqrt{v} d v+C \\\\& \Rightarrow y \sqrt{\log x}=\frac{2 v^{3 / 2}}{3}+C \\\\& \Rightarrow y \sqrt{\log x}=\frac{2}{3}(\log x)^{3 / 2}+C\end{aligned} am p ; y log x = ∫ x log x d x am p ; log x = v ⇒ x 1 d x = d v am p ; ⇒ y log x = ∫ v d v + C am p ; ⇒ y log x = 3 2 v 3/2 + C am p ; ⇒ y log x = 3 2 ( log x ) 3/2 + C Now, this curve passes through
( e , 4 3 ) \left(e, \frac{4}{3}\right) ( e , 3 4 ) and
( e 4 , α ) \left(e^4, \alpha\right) ( e 4 , α ) a m p ; ∴ 4 3 log e = 2 3 ( log e ) 3 / 2 + C a m p ; ⇒ C = 4 3 − 2 3 = 2 3 \begin{aligned}& \therefore \frac{4}{3} \sqrt{\log e}=\frac{2}{3}(\log e)^{3 / 2}+C \\\\& \Rightarrow C=\frac{4}{3}-\frac{2}{3}=\frac{2}{3}\end{aligned} am p ; ∴ 3 4 log e = 3 2 ( log e ) 3/2 + C am p ; ⇒ C = 3 4 − 3 2 = 3 2 Also,
α log e 4 = 2 3 ( log e 4 ) 3 / 2 + 2 3 \alpha \sqrt{\log e^4}=\frac{2}{3}\left(\log e^4\right)^{3 / 2}+\frac{2}{3} α log e 4 = 3 2 ( log e 4 ) 3/2 + 3 2 a m p ; ⇒ 2 α = 2 3 × ( 4 ) 3 / 2 + 2 3 = 16 3 + 2 3 = 18 3 a m p ; ⇒ α = 3 \begin{aligned}& \Rightarrow 2 \alpha=\frac{2}{3} \times(4)^{3 / 2}+\frac{2}{3}=\frac{16}{3}+\frac{2}{3}=\frac{18}{3} \\\\& \Rightarrow \alpha=3\end{aligned} am p ; ⇒ 2 α = 3 2 × ( 4 ) 3/2 + 3 2 = 3 16 + 3 2 = 3 18 am p ; ⇒ α = 3