JEE Advanced 2021 — Paper 2 — Question 20

The given differential equation is $\frac{dy}{dx} + \alpha y = x e^{\beta x}$. Multiply both sides by integrating factor $e^{\alpha x}$: $\frac{d}{dx}(e^{\alpha x} y) = x e^{(\alpha+\beta)x}$. Case 1: $\alpha + \beta eq 0$. Integrate: $e^{\alpha x} y = \int x e^{(\alpha+\beta)x} dx = \frac{x e^{(\alpha+\beta)x}}{\alpha+\beta} - \frac{e^{(\alpha+\beta)x}}{(\alpha+\beta)^2} + C$.

Hence $y = \frac{x e^{\beta x}}{\alpha+\beta} - \frac{e^{\beta x}}{(\alpha+\beta)^2} + C e^{-\alpha x}$.

Using $y(1)=1$ gives $C = e^{\alpha}\left(1 - \frac{1 e^{\beta}}{\alpha+\beta} + \frac{e^{\beta}}{(\alpha+\beta)^2}\right)$.

For $\alpha=1, \beta=1$, $y = \frac{x e^x}{2} - \frac{e^x}{4} + \left(e - \frac{e^2}{4}\right)e^{-x}$. Case 2: $\alpha + \beta = 0$.

Then $\frac{d}{dx}(e^{\alpha x} y) = x$, so $e^{\alpha x} y = \frac{x^2}{2} + C$, i.e. $y = e^{-\alpha x}\left(\frac{x^2}{2} + C\right) = e^{\beta x} \left(\frac{x^2}{2}+C\right)$.

Using $y(1)=1$ gives $C = e^{-\beta} - \frac12$.

For $\beta = -1$, $y = \frac{x^2}{2} e^{-x} + \left(e - \frac12\right)e^{-x}$. The two derived solutions match options C and A respectively.