The solution to the differential equation y l | vedclass

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

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$x \ln y = 1$

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(A) Given the differential equation: $y \ln y + x \frac{dy}{dx} = 0$. Rearranging the terms to separate the variables: $\frac{dx}{x} + \frac{dy}{y \ln y} = 0$. Integrating both sides: $\int \frac{dx}{x} + \int \frac{dy}{y \ln y} = C$. Let $u = \ln y$,then $du = \frac{1}{y} dy$. The integral becomes: $\ln |x| + \ln |\ln y| = C$. This simplifies to: $\ln |x \ln y| = C$,or $x \ln y = k$ where $k = e^C$. Using the initial condition $y(1) = e$: $1 \cdot \ln(e) = k \Rightarrow 1 \cdot 1 = k \Rightarrow k = 1$. Thus,the solution is $x \ln y = 1$.

The solution to the differential equation $y \ln y + xy' = 0,$ where $y(1) = e,$ is

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