General solution of the differential equation | vedclass

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Question

(A) 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 \l

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given differential equation: $x \cos y \,dy = (x e^x \log x + e^x) dx$ Divide both sides by $x$ (assuming $x 
eq 0$): $\cos y \,dy = \left(e^x \log x + \frac{e^x}{x}ight) dx$ Integrating both sides: $\int \cos y \,dy = \int e^x \left(\log x + \frac{1}{x}ight) dx$ Using the standard integral formula $\int e^x \{f(x) + f'(x)\} dx = e^x f(x) + C$, where $f(x) = \log x$ and $f'(x) = \frac{1}{x}$: $\sin y = e^x \log x + C$

General solution of the differential equation $x \cos y \,dy = (x e^x \log x + e^x) dx$ is (where $C$ is a constant of integration.)

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