The solution of the differential equation x c | vedclass

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Question

(C) Given differential equation: $x \cos y \, dy = (x e^x \log x + e^x) \, dx$ Divide both sides by $x$ (assuming $x 
eq 0$): $\cos y \, dy = \left(

Vedclass · Accepted Answer

$\sin y = e^x \log x + c$

Vedclass · Answer

(C) Given differential equation: $x \cos y \, dy = (x e^x \log x + e^x) \, dx$ Divide both sides by $x$ (assuming $x 
eq 0$): $\cos y \, dy = \left( e^x \log x + \frac{e^x}{x} ight) \, dx$ Integrate both sides: $\int \cos y \, dy = \int (e^x \log x + \frac{e^x}{x}) \, dx$ We know that $\frac{d}{dx} (e^x \log x) = e^x \log x + e^x \cdot \frac{1}{x} = e^x \log x + \frac{e^x}{x}$. Therefore,$\int (e^x \log x + \frac{e^x}{x}) \, dx = e^x \log x + c$. Thus,the solution is $\sin y = e^x \log x + c$.

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

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