The general solution of the differential equa | vedclass

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Question

(B) Given the differential equation: $x \cos y \,dy = (x e^x \log x + e^x) dx$. Dividing both sides by $x$ (assuming $x 
eq 0$): $\cos y \,dy = \frac

Vedclass · Accepted Answer

$\sin y = e^x \log x + c$,where $c$ is a constant of integration.

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(B) Given the differential equation: $x \cos y \,dy = (x e^x \log x + e^x) dx$. Dividing both sides by $x$ (assuming $x 
eq 0$): $\cos y \,dy = \frac{x e^x \log x + e^x}{x} dx$ $\cos y \,dy = (e^x \log x + \frac{e^x}{x}) dx$ We observe that the right side is the derivative of $e^x \log x$ with respect to $x$. Using the product rule: $\frac{d}{dx}(e^x \log x) = e^x \log x + e^x \cdot \frac{1}{x} = e^x \log x + \frac{e^x}{x}$. Integrating both sides: $\int \cos y \,dy = \int (e^x \log x + \frac{e^x}{x}) dx$ $\sin y = e^x \log x + c$,where $c$ is a constant of integration.

The general solution of the differential equation $x \cos y \,dy = (x e^x \log x + e^x) dx$ is given by

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