For Any Real Numbers α and β, Let y(α,β)(x) Be the Solution of dy/dx + αy = xe^(βx), y(1) = 1. If S = {y(α,β)(x) : α, β ∈ R}, Then Which of the Following Functions Belong to S

JEE Advanced 2021, Paper 2. Previous Year Question.

Problem

For any real numbers $\alpha$ and $\beta$, let $y_{\alpha,\beta}(x)$, $x \in \mathbb{R}$, be the solution of the differential equation

$$\frac{dy}{dx} + \alpha y = xe^{\beta x}, \qquad y(1) = 1$$

Let $S = \{y_{\alpha,\beta}(x) : \alpha, \beta \in \mathbb{R}\}$. Then which of the following functions belong(s) to the set $S$?

Key insight. Solving the differential equation for general $\alpha,\beta$ splits naturally into two cases: $\alpha+\beta \neq 0$ (where the integral is straightforward) and the special case $\alpha+\beta=0$ (where a term that looked fine for generic parameters would blow up, forcing a different computation). Missing the second case is the single most common way to lose marks on this question, since it’s a “multiple correct” problem and each case matches a different option.

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Approach

This is a first-order linear differential equation in $y$ with integrating factor $e^{\alpha x}$, but the parameter $\beta$ appears in the exponent on the right-hand side, and after integrating, the result naturally divides by $(\alpha+\beta)$ — a step that’s only valid when $\alpha+\beta \neq 0$. So the problem has to be split into two separate cases, each giving a different family of solutions, and each of those families is checked against the four given options separately.

Solution

Step 1 — Solve for the case α + β ≠ 0

The integrating factor is $e^{\int\alpha\,dx} = e^{\alpha x}$. Multiplying through:

$$\frac{d}{dx}\left(y\,e^{\alpha x}\right) = xe^{\beta x}\cdot e^{\alpha x} = xe^{(\alpha+\beta)x}$$

Integrating the right-hand side by parts (with $u=x$, $dv=e^{(\alpha+\beta)x}dx$):

$$y\,e^{\alpha x} = \frac{x\,e^{(\alpha+\beta)x}}{\alpha+\beta} – \frac{e^{(\alpha+\beta)x}}{(\alpha+\beta)^2} + C$$

so

$$y = \frac{x\,e^{\beta x}}{\alpha+\beta} – \frac{e^{\beta x}}{(\alpha+\beta)^2} + Ce^{-\alpha x}$$

Using $y(1)=1$ to fix $C$, then substituting back, gives the general solution in this case. This is valid provided $\alpha+\beta \neq 0$.

Step 2 — Try α = β = 1 and match against the options

Setting $\alpha=\beta=1$ (so $\alpha+\beta=2\neq0$) and applying $y(1)=1$ to solve for $C$, then simplifying by factoring $e^{x/2}$ (obtained by combining $e^{\beta x}=e^x$ and $e^{-\alpha x}=e^{-x}$, then re-expressing around $x/2$ to match the option format) produces

$$y = \frac{e^x}{2}\left(x-\frac12\right) + \left(e – \frac{e^2}{4}\right)e^{-x}$$

which is exactly option C. So option C belongs to $S$ (with $\alpha=\beta=1$).

Step 3 — Handle the special case α + β = 0

When $\alpha+\beta=0$ (i.e. $\beta=-\alpha$), the term $\dfrac{1}{\alpha+\beta}$ from Step 1 is undefined, so this case must be solved separately. The differential equation becomes $\dfrac{dy}{dx}+\alpha y = xe^{-\alpha x}$, and multiplying by the integrating factor $e^{\alpha x}$:

$$\frac{d}{dx}\left(y\,e^{\alpha x}\right) = xe^{-\alpha x}\cdot e^{\alpha x} = x$$

Integrating directly (no division by $\alpha+\beta$ needed here):

$$y\,e^{\alpha x} = \frac{x^2}{2} + C \implies y = \frac{x^2}{2}e^{-\alpha x} + Ce^{-\alpha x}$$

Step 4 — Try α = 1, β = −1 and match against the options

With $\alpha=1$, apply $y(1)=1$: $1 = \dfrac12 e^{-1} + Ce^{-1} \implies C = e-\dfrac12$. Substituting back:

$$y = \frac{x^2}{2}e^{-x} + \left(e-\frac12\right)e^{-x}$$

which is exactly option A. So option A also belongs to $S$ (with $\alpha=1,\ \beta=-1$).

Step 5 — Check options B and D

Comparing the same computations against options B and D shows their constant terms and signs don’t match either case’s algebra (the sign on the $\dfrac{x^2}{2}e^{-x}$ term in B is flipped, and the constant in D doesn’t come out of either case’s substitution), so neither is achievable for any real $\alpha,\beta$.

Answer

Common mistakes

• Only solving the α + β ≠ 0 case and stopping there. This “multiple correct” question is built precisely so that the general-case formula misses one of the two correct options — the $\alpha+\beta=0$ branch has to be worked out as its own separate differential equation.
• Forgetting that $\alpha+\beta=0$ makes $\dfrac{1}{\alpha+\beta}$ meaningless. It’s tempting to substitute $\beta=-\alpha$ directly into the Step 1 formula, but that formula was only ever valid because we divided by $\alpha+\beta$ along the way — so it has to be re-derived from scratch for this case.
• Picking the wrong trial values for α and β. Since $S$ is defined over all real $\alpha,\beta$, matching an option means finding some valid pair — the options’ specific numerical constants (like $e-\frac12$ or $e-\frac{e^2}{4}$) are the clue for which simple values of $\alpha,\beta$ (like $1$ and $\pm1$) to try first.

Practise next

• For the same family $y_{\alpha,\beta}(x)$ solving $\dfrac{dy}{dx}+\alpha y = xe^{\beta x}$, $y(1)=1$, determine whether $f(x) = \dfrac{x^2}{2}e^{x} + \left(e^{-1}-\dfrac12\right)e^{x}$ can belong to $S$ for some choice of $\alpha,\beta$.