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Question

(A) Given the differential equation: $y(1+\log x) \frac{dx}{dy} - x \log x = 0$ Rearranging the terms to separate the variables: $y(1+\log x) \frac{dx

Vedclass · Accepted Answer

$y = ex \log x$

Vedclass · Answer

(A) Given the differential equation: $y(1+\log x) \frac{dx}{dy} - x \log x = 0$ Rearranging the terms to separate the variables: $y(1+\log x) \frac{dx}{dy} = x \log x$ $\frac{1+\log x}{x \log x} dx = \frac{dy}{y}$ Integrating both sides: $\int \frac{1+\log x}{x \log x} dx = \int \frac{dy}{y}$ Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes $\int \frac{1+u}{u} du = \int (\frac{1}{u} + 1) du = \log u + u$. Substituting back $u = \log x$: $\log(\log x) + \log x = \log y + C$ Alternatively,using substitution $v = x \log x$,$dv = (1 + \log x) dx$: $\int \frac{dv}{v} = \int \frac{dy}{y}$ $\log(x \log x) = \log y + \log C$ $\log(x \log x) = \log(Cy)$ $x \log x = Cy$ Given $x = e$ and $y = e^2$: $e \log e = C(e^2)$ $e(1) = Ce^2 \Rightarrow C = \frac{1}{e}$ Substituting $C$ back into the equation: $x \log x = \frac{y}{e}$ $y = ex \log x$

The particular solution of the differential equation $y(1+\log x) \frac{dx}{dy} - x \log x = 0$ given that $y = e^2$ when $x = e$ is:

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