Question

The solution of \( (y \cos y + \sin y) \, dy = (2x \log x + x) \, dx \) is:

• A. \( y \sin x = x^2 \log x + C \)
• B. \( y \sin y = x \log x + C \)
• C. \( y \sin y = x^2 \log x + C \)
• D. \( \sin x = x^2 \log x + C \)
• E. \( y \sin x = x \log x + C \)

Hint:

For separable differential equations, separate the variables and integrate each side separately. Use integration by parts where needed for terms involving products like \( x \log x \) or \( y \cos y \).

Answer 1

The Correct Option is C

We are given the differential equation: \[ (y \cos y + \sin y) \, dy = (2x \log x + x) \, dx \] First, let's separate the variables \( y \) and \( x \) on each side. Rearranging the equation gives: \[ \frac{dy}{dx} = \frac{2x \log x + x}{y \cos y + \sin y} \] This is a separable differential equation, so let's proceed by integrating both sides: \[ \int (y \cos y + \sin y) \, dy = \int (2x \log x + x) \, dx \] The integral on the left-hand side: \[ \int (y \cos y + \sin y) \, dy \] can be solved by first separating the terms: \[ \int y \cos y \, dy + \int \sin y \, dy \] We use integration by parts to solve \( \int y \cos y \, dy \), and the integral of \( \sin y \) is straightforward: \[ \int \sin y \, dy = -\cos y \] For \( \int y \cos y \, dy \), use the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \] Let \( u = y \) and \( dv = \cos y \, dy \). Then: \[ du = dy, \quad v = \sin y \] \[ \int y \cos y \, dy = y \sin y - \int \sin y \, dy = y \sin y + \cos y \] Thus, the left-hand side integral becomes: \[ y \sin y + \cos y \] Now, for the right-hand side: \[ \int (2x \log x + x) \, dx \] The integral of \( x \) is straightforward: \[ \int x \, dx = \frac{x^2}{2} \] For \( \int 2x \log x \, dx \), use integration by parts. Let \( u = \log x \) and \( dv = 2x \, dx \), then: \[ du = \frac{1}{x} \, dx, \quad v = x^2 \] Thus: \[ \int 2x \log x \, dx = x^2 \log x - \int x \, dx = x^2 \log x - \frac{x^2}{2} \] Therefore, the right-hand side integral becomes: \[ x^2 \log x - \frac{x^2}{2} \] Putting it all together: \[ y \sin y + \cos y = x^2 \log x - \frac{x^2}{2} + C \] Thus, the solution to the differential equation is: \[ y \sin y = x^2 \log x + C \] Thus, the correct answer is option (C), \( y \sin y = x^2 \log x + C \).