MHT CET · Maths · Differential Equations
The particular solution of the differential equation \(y\left(\frac{\mathrm{d} x}{\mathrm{~d} y}\right)=x \log x\) at \(x=\mathrm{e}\) and \(\mathrm{y}=1\) is
- A \(\mathrm{e}^{x y}=2\)
- B \(x=\mathrm{e}^{\mathrm{y}}\)
- C \(x\mathrm{y}=2\)
- D \(\log x=2 y\)
Answer & Solution
Correct Answer
(B) \(x=\mathrm{e}^{\mathrm{y}}\)
Step-by-step Solution
Detailed explanation
(D)
\(y\left(\frac{d x}{d y}\right)=x \cdot \log x\)
\(\therefore \int \frac{1}{x \cdot \log x} d x=\int \frac{1}{y} d y\)
\(\therefore \log |\log x|=\log y+\log c\)
We have \(x=e\) and \(y=1\)
\(\therefore \log |\log e|=\log 1+\log c \Rightarrow \log c=0\)
\(\therefore \log |\log x|=\log y \Rightarrow \log x=y \Rightarrow x=e^{y}\)
\(y\left(\frac{d x}{d y}\right)=x \cdot \log x\)
\(\therefore \int \frac{1}{x \cdot \log x} d x=\int \frac{1}{y} d y\)
\(\therefore \log |\log x|=\log y+\log c\)
We have \(x=e\) and \(y=1\)
\(\therefore \log |\log e|=\log 1+\log c \Rightarrow \log c=0\)
\(\therefore \log |\log x|=\log y \Rightarrow \log x=y \Rightarrow x=e^{y}\)
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