The solution of (xy cos xy + sin xy)dx + x^2 | vedclass

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Question

(A) Given differential equation: $(xy \cos xy + \sin xy)dx + x^2 \cos xy \, dy = 0$ Rearranging the terms: $xy \cos xy \, dx + x^2 \cos xy \, dy + \si

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$x \sin (xy) = k$

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(A) Given differential equation: $(xy \cos xy + \sin xy)dx + x^2 \cos xy \, dy = 0$ Rearranging the terms: $xy \cos xy \, dx + x^2 \cos xy \, dy + \sin xy \, dx = 0$ Factor out $x \cos xy$ from the first two terms: $x \cos xy (y \, dx + x \, dy) + \sin xy \, dx = 0$ We know that $d(xy) = y \, dx + x \, dy$. Substituting this: $x \cos xy \, d(xy) + \sin xy \, dx = 0$ Divide the entire equation by $x \sin xy$: $\cot (xy) \, d(xy) + \frac{dx}{x} = 0$ Integrating both sides: $\int \cot (xy) \, d(xy) + \int \frac{1}{x} \, dx = \int 0 \, dx$ $\ln |\sin (xy)| + \ln |x| = C$ Using the property $\ln a + \ln b = \ln (ab)$: $\ln |x \sin (xy)| = C$ Taking the exponent on both sides: $x \sin (xy) = e^C = k$ Thus,the solution is $x \sin (xy) = k$.

The solution of $(xy \cos xy + \sin xy)dx + x^2 \cos xy \, dy = 0$ is

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