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Integrating factor of $(x\, log_ex)\frac{dy}{dx}+y = 2log_e x $ is : |
x $e^x$ $log_ex$ $log_e(log_ex)$ |
$log_ex$ |
The correct answer is option (3) → $\log_ex$ dividing eq. by $x\log x$ so $\frac{dy}{dx}+\frac{y}{x\log x}=\frac{2}{x}$ $I.F.=e^{\int\frac{1}{x\log x}}dx=e^{\log\log x}=\log x$ |
