Kerala KEAM 5 June 2024 Solved Paper

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), $y \sin y = x^2 \log x + C$. Quick Tip: 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 $y \cos y$.