MHT CET · Maths · Differential Equations
The general solution of the differential equation \(x \cos y \mathrm{~d} y=\left(x \mathrm{e}^x \log x+\mathrm{e}^x\right) \mathrm{d} x\) is given by
- A \(\sin y=\mathrm{e}^x+\operatorname{cog} x\), where c is a constant of integration.
- B \(\sin y=\mathrm{e}^x \log x+\mathrm{c}\), where c is a constant of integration.
- C \(\mathrm{e}^x \sin y=\log x+\mathrm{c}\), where c is a constant of integration,
- D \(\sin y=\mathrm{ce}^x+\log x\), where c is a constant of integration.
Answer & Solution
Correct Answer
(B) \(\sin y=\mathrm{e}^x \log x+\mathrm{c}\), where c is a constant of integration.
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& x \cos y \mathrm{~d} y=\left(x \mathrm{e}^x \log x+\mathrm{e}^x\right) \mathrm{d} x \\
& \Rightarrow \cos y \mathrm{~d} y=\mathrm{e}^x\left(\log x+\frac{1}{x}\right) \mathrm{d} x
\end{aligned}\)
Integrating on both sides, we get
\(\sin y=\mathrm{e}^x \log x+\mathrm{c}\)
& x \cos y \mathrm{~d} y=\left(x \mathrm{e}^x \log x+\mathrm{e}^x\right) \mathrm{d} x \\
& \Rightarrow \cos y \mathrm{~d} y=\mathrm{e}^x\left(\log x+\frac{1}{x}\right) \mathrm{d} x
\end{aligned}\)
Integrating on both sides, we get
\(\sin y=\mathrm{e}^x \log x+\mathrm{c}\)
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