BITSAT 2024 Shift 1 Paper

To solve the given differential equation$(x+1)\;\frac{dy}{dx}-y=e^{3x}(x+1)^2$we start by rewriting it in the standard linear form:$\frac{dy}{dx}-\frac{y}{x+1}=e^{3x}(x+1)$Here, it's a first-order linear differential equation of the form$\frac{dy}{dx}+P(x)y=Q(x)$where $P(x)=-\frac{1}{x+1}$ and $Q(x)=e^{3x}(x+1)$.Step 1: Calculate the Integrating Factor (IF)The integrating factor is given by:$IF = e^{\int P(x) dx} = e^{\int -\frac{1}{x+1} dx} = e^{-\ln(x+1)} = \frac{1}{x+1}$Step 2: Determine the SolutionUsing the integrating factor, the solution of the differential equation is:$y \cdot IF = \int Q(x) \cdot IF \, dx$Substituting the known values:$y \cdot \frac{1}{x+1} = \int e^{3x}(x+1) \cdot \frac{1}{x+1} \, dx$This simplifies to:$\frac{y}{x+1} = \frac{e^{3x}}{3} + C'$Therefore, the complete solution is: $\frac{3y}{x+1} = e^{3x} + C'$where $C' = \frac{C}{3}$ representing an integration constant.