An integrating factor of the differential equation \(x \frac{d y}{d x}+y \log x=x^x x^{-\frac{1}{2}},(x>0)\)…
- \(x^{\log x}\)
- \((\sqrt{x})^{\log x}\)
- \((\sqrt{\mathrm{e}})^{(\log x)^2}\)
- \(\mathrm{e}^{\mathrm{x}^2}\)
Solution
\(\begin{aligned}
& x \frac{d y}{d x}+y \log x=e^x x^{(-1 / 2) \log x} \\
& \text {or } \frac{d y}{d x}+y \cdot \frac{1}{x} \log x=e^x x^{-(1 / 2) \log x} \\
& \text {Here, } P=\frac{1}{x} \log x \text { and } Q=e^x x^{-(1 / 2) \log x} \\
& \therefore I F=e^{\int \frac{1}{x} \log x d x} \\
& \text {Put } \log x=t \\
& \frac{1}{x} d x=d t \\
& \text {IF }=e^{\int t d t} \\
& \text {IF }=e^{\frac{t^2}{2}} \\
& \frac{(\log x)^2}{2}=(\sqrt{e})^{(\log x)^2} .
\end{aligned}\)
Asked in: MHT CET 2020 (19 Oct Shift 2)