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Question

(A) Given the differential equation: $\frac{dy}{dx} = \frac{\sin y + e^x}{\ln y - x \cos y}$.Rearranging the terms,we get: $(\ln y - x \cos y) dy = (\

Vedclass · Accepted Answer

$y(\ln y - 1) = e^x + x \sin y + C$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = \frac{\sin y + e^x}{\ln y - x \cos y}$.Rearranging the terms,we get: $(\ln y - x \cos y) dy = (\sin y + e^x) dx$.This can be rewritten as: $\ln y \, dy - x \cos y \, dy = \sin y \, dx + e^x \, dx$.Grouping the terms: $\ln y \, dy - e^x \, dx = x \cos y \, dy + \sin y \, dx$.Recognizing the product rule on the right side,$d(x \sin y) = x \cos y \, dy + \sin y \, dx$.So,the equation becomes: $\ln y \, dy - e^x \, dx = d(x \sin y)$.Integrating both sides: $\int \ln y \, dy - \int e^x \, dx = \int d(x \sin y)$.Using the integral formula $\int \ln y \, dy = y \ln y - y$,we get:$(y \ln y - y) - e^x = x \sin y + C$.Rearranging gives: $y(\ln y - 1) = e^x + x \sin y + C$.

The solution of the differential equation $\frac{dy}{dx} = \frac{\sin y + e^x}{\ln y - x \cos y}$ is:

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