The integrating factor of the differential eq | vedclass

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(D) Given the differential equation: $3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x$. Dividing both sides by $3 x \log _{e} x$, we get: $\frac{d y}{

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$(\log _{e} x)^{1 / 3}$

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(D) Given the differential equation: $3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x$. Dividing both sides by $3 x \log _{e} x$, we get: $\frac{d y}{d x} + \frac{1}{3 x \log _{e} x} y = \frac{2}{3 x}$. This is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$, where $P = \frac{1}{3 x \log _{e} x}$. The integrating factor $(IF)$ is given by $e^{\int P d x}$. $IF = e^{\int \frac{1}{3 x \log _{e} x} d x}$. Let $t = \log _{e} x$, then $d t = \frac{1}{x} d x$. $IF = e^{\frac{1}{3} \int \frac{1}{t} d t} = e^{\frac{1}{3} \log _{e} t} = e^{\log _{e} (t^{1/3})} = t^{1/3}$. Substituting back $t = \log _{e} x$, we get $IF = (\log _{e} x)^{1/3}$.

The integrating factor of the differential equation $3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x$ is given by

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