KCET · Maths · Differential Equations
If \(y(x)\) is the solution of differential equation \(x \log x \frac{d y}{d x}+y=2 x \log x, y(e)\) is equal to
- A \(e\)
- B 0
- C 2
- D \(2 e\)
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
\(x \log x \frac{d y}{d x}+y=2 x \log x \Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=2\)
\[
\begin{aligned}
& \mathrm{IF}=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x \\
& y(\log x)=\int(2)(\log x) \cdot d x+C \\
& \Rightarrow \quad y \log x=2 \int \log x d x+C \\
& \Rightarrow \quad y \log x=2[x \log x-x]+C
\end{aligned}
\]
Putting \(x=1\), we get
\[
y \cdot 0=2[1 \times 0-1]+C \Rightarrow C=2
\]
Putting \(C=2\) in Eq. (i), we get
\[
y \log x=2[x \log x-x]+2
\]
At
\[
\begin{aligned}
x & =e \Rightarrow y \cdot 1=2(e \times 1-e)+2 \\
y & =0+2=2 \\
y(e) & =2
\end{aligned}
\]
\[
\begin{aligned}
& \mathrm{IF}=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x \\
& y(\log x)=\int(2)(\log x) \cdot d x+C \\
& \Rightarrow \quad y \log x=2 \int \log x d x+C \\
& \Rightarrow \quad y \log x=2[x \log x-x]+C
\end{aligned}
\]
Putting \(x=1\), we get
\[
y \cdot 0=2[1 \times 0-1]+C \Rightarrow C=2
\]
Putting \(C=2\) in Eq. (i), we get
\[
y \log x=2[x \log x-x]+2
\]
At
\[
\begin{aligned}
x & =e \Rightarrow y \cdot 1=2(e \times 1-e)+2 \\
y & =0+2=2 \\
y(e) & =2
\end{aligned}
\]
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