MHTCET 22 Apr 2025 Shift 1 Paper

Concept: The equation is a linear differential equation of the form $y' + P y = Q$, solved using the integrating factor method.Explanation:Given $y+\frac{d}{d x}(x y)=x(\sin x+\log x)$.Expanding, $\frac{d}{d x}(x y)=x \frac{d y}{d x}+y$.So, $y+x \frac{d y}{d x}+y=x(\sin x+\log x)$.Simplifying, $x \frac{d y}{d x}+2y=x(\sin x+\log x)$.Dividing by $x$, we get $\frac{d y}{d x}+\frac{2}{x}y=\sin x+\log x$.Here, $P=\frac{2}{x}$ and $Q=\sin x+\log x$.Integrating factor, $\text{I.F.}=e^{\int P\,dx}=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^{2}$.The solution is $y\cdot x^{2}=\int x^{2}(\sin x+\log x)\,dx+c$.Now, $\int x^{2}\sin x\,dx=-x^{2}\cos x+2x\sin x+2\cos x$.And $\int x^{2}\log x\,dx=\frac{x^{3}}{3}\log x-\frac{x^{3}}{9}$.Therefore, $y x^{2}=-x^{2}\cos x+2x\sin x+2\cos x+\frac{x^{3}}{3}\log x-\frac{x^{3}}{9}+c$.Dividing by $x^{2}$ gives the required solution.Answer:$y=-\cos x+\frac{2\sin x}{x}+\frac{2}{x^{2}}\cos x+\frac{x}{3}\log x-\frac{x}{9}+\frac{c}{x^{2}}$, where $c$ is the constant of integration.This matches Option C.