MHT CET · Maths · Differential Equations
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 (where \(C\) is a constant of integration.)
- A \(\sin y=\mathrm{e}^x \log x+C\)
- B \(\sin y=\mathrm{e}^x+C \log x\)
- C \(\sin y=C \mathrm{e}^x+\log x\)
- D \(\mathrm{e}^x \sin y=\log x+C\)
Answer & Solution
Correct Answer
(A) \(\sin y=\mathrm{e}^x \log x+C\)
Step-by-step Solution
Detailed explanation
\(x \cos y \mathrm{~d} y=\left(x \mathrm{e}^x \log x+\mathrm{e}^x\right) \mathrm{d} x\)
\( \Rightarrow \int \cos y \mathrm{~d} y=\int \mathrm{e}^x\left\{\log x+\frac{1}{x}\right\} \mathrm{d} x\)
\(\Rightarrow \sin y=\mathrm{e}^x \log x+C[\because \int \mathrm{e}^x\left\{f(x)+f^{\prime}(x)\right\} \mathrm{d} x~=\) \(\mathrm{e}^x f(x)+C]\)
\( \Rightarrow \int \cos y \mathrm{~d} y=\int \mathrm{e}^x\left\{\log x+\frac{1}{x}\right\} \mathrm{d} x\)
\(\Rightarrow \sin y=\mathrm{e}^x \log x+C[\because \int \mathrm{e}^x\left\{f(x)+f^{\prime}(x)\right\} \mathrm{d} x~=\) \(\mathrm{e}^x f(x)+C]\)
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