MHT CET · Maths · Differential Equations
The general solution of the differential equation
\(
y(1+\log x)\left(\frac{d x}{d y}\right)
\) \(-x \log x=0\) is
- A \(\mathrm{y}(1+\log \mathrm{x})=\mathrm{c}\)
- B \(\mathrm{x} \log \mathrm{x}=\mathrm{yc}\)
- C \(x \log x=y+c\)
- D \(\log \mathrm{x}-\mathrm{y}=\mathrm{c}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{x} \log \mathrm{x}=\mathrm{yc}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& y(1+\log x)\left(\frac{d y}{d x}\right)-x \log x=0 \\
& \therefore \int \frac{(1+\log x)}{x \log x} d x=\int \frac{d y}{y} \\
& \int \frac{d x}{x \log x}+\int \frac{d x}{x}=\int \frac{d y}{y} \\
& \therefore \log (\log x)+\log x=\log y+\log c \\
& \therefore \log [x \log x]=\log (y, c)
\Rightarrow x \log x=y c
\end{aligned}
\)
\begin{aligned}
& y(1+\log x)\left(\frac{d y}{d x}\right)-x \log x=0 \\
& \therefore \int \frac{(1+\log x)}{x \log x} d x=\int \frac{d y}{y} \\
& \int \frac{d x}{x \log x}+\int \frac{d x}{x}=\int \frac{d y}{y} \\
& \therefore \log (\log x)+\log x=\log y+\log c \\
& \therefore \log [x \log x]=\log (y, c)
\Rightarrow x \log x=y c
\end{aligned}
\)
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