The integrating factor of the differential eq | vedclass

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Question

(D) Given the differential equation: $\frac{dy}{dx}(x \log x) + y = 4 \log x$. Divide the entire equation by $(x \log x)$ to bring it into the standar

Vedclass · Accepted Answer

$\log x$

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(D) Given the differential equation: $\frac{dy}{dx}(x \log x) + y = 4 \log x$. Divide the entire equation by $(x \log x)$ to bring it into the standard form $\frac{dy}{dx} + Py = Q$: $\frac{dy}{dx} + \frac{1}{x \log x} y = \frac{4 \log x}{x \log x} = \frac{4}{x}$. Here,$P = \frac{1}{x \log x}$. The integrating factor ($I$.$F$.) is given by $e^{\int P dx}$: $I.F. = e^{\int \frac{1}{x \log x} dx}$. Let $u = \log x$,then $du = \frac{1}{x} dx$. So,$\int \frac{1}{x \log x} dx = \int \frac{1}{u} du = \log u = \log(\log x)$. Therefore,$I.F. = e^{\log(\log x)} = \log x$.

The integrating factor of the differential equation $\frac{dy}{dx}(x \log x) + y = 4 \log x$ is

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