The particular solution of the differential e | vedclass

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Question

(B) Given the differential equation: $y \frac{dx}{dy} = x \log x$ Separate the variables: $\frac{dx}{x \log x} = \frac{dy}{y}$ Integrate both sides: $

Vedclass · Accepted Answer

$x = e^y$

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(B) Given the differential equation: $y \frac{dx}{dy} = x \log x$ Separate the variables: $\frac{dx}{x \log x} = \frac{dy}{y}$ Integrate both sides: $\int \frac{1}{x \log x} dx = \int \frac{1}{y} dy$ Let $u = \log x$,then $du = \frac{1}{x} dx$. The integral becomes $\int \frac{1}{u} du = \log |u| = \log |\log x|$. So,$\log |\log x| = \log |y| + C$. Given $x = e$ and $y = 1$: $\log |\log e| = \log |1| + C \Rightarrow \log |1| = 0 + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$. Thus,$\log |\log x| = \log |y|$. Taking the exponential of both sides: $\log x = y$,which implies $x = e^y$.

The particular solution of the differential equation $y(\frac{dx}{dy}) = x \log x$ at $x = e$ and $y = 1$ is

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