Q120 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics
If $x \log x \frac{d y}{d x}+y=\log x^2$ and $y(e)=0$, then $y\left(e^2\right)=$
- A0
- B1
- C$\frac{1}{2}$
- Correct answerD$\frac{3}{2}$
Solution
We have,
$$
\begin{aligned}
& x \log x \frac{d y}{d x}+y=\log x^2 \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2 \log x}{x \log x} \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \\
& \mathrm{IF}=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)} \\
& \mathrm{IF}=\log x \\
&
\end{aligned}
$$
Solution of different equation is
$$
\begin{array}{rlrl}
& & y \log x & =\int \frac{2 \log x}{x} d x+c \\
y \log x & =(\log x)^2+c \\
\text { put } x & =e, y=0 \text { we get } \\
0 & =1+c \Rightarrow c=-1 \\
\therefore \quad & y \log x & =(\log x)^2-1 \\
\text { put } & & =e^2 \\
y \log e^2 & =\left(\log e^2\right)^2-1 \\
\Rightarrow & & 2 y & =4-1 \\
\therefore & & y & =\frac{3}{2} \\
& & y\left(e^2\right) & =\frac{3}{2}
\end{array}
$$
Hence, the correct answer is option: 'D'.