For any real numbers α and β, let yα,β(x), x ∈ ℝ, be the solution of the differential equation dy/dx

Q: Wait, why do we split the problem into two cases? Can't we just use one formula?

That is a great question! The integral $\int x e^{kx} dx$ behaves differently based on the value of $k$. If $k = 0$, the term $e^{kx}$ becomes $1$, and the integral is simply $\int x dx = \frac{x^2}{2} + C$. If $k \neq 0$, we must use Integration by Parts. Since $\alpha$ and $\beta$ are arbitrary real numbers, we cannot assume $\alpha + \beta$ is non-zero. Thus, we must analyze both scenarios to find all possible functions in set $S$.

Q: How do I know which values of $\alpha$ and $\beta$ to test for the options?

Don't guess! Instead, look at the structure of the options. For example, Option A has a term with $x^2$, which suggests it came from the Case 1 ($\alpha + \beta = 0$) solution. Option C has terms with $x e^x$ and $e^x$, which suggests it came from the Case 2 ($\alpha + \beta \neq 0$) solution. By matching the form of the option to our derived general solution, you can deduce the specific values of $\alpha$ and $\beta$ required.

Q: Is there a shortcut to avoid Integration by Parts for Case 2?

While there is no 'shortcut' that bypasses the math, you can use the method of undetermined coefficients if you are comfortable with it. However, for JEE Advanced, mastering Integration by Parts is essential. The integral $\int x e^{kx} dx$ is a standard form. Just remember: $u=x$ and $dv=e^{kx} dx$. With practice, this becomes second nature!

Question

For any real numbers $\alpha$ and $\beta$, let $y_{\alpha, \beta}(x), x \in \mathbb{R}$, be the solution of the differential equation $\frac{dy}{dx} + \alpha y = x e^{\beta x}, y(1) = 1$. Let $S = \{y_{\alpha, \beta}(x) : \alpha, \beta \in \mathbb{R}\}$. Then which of the following functions belong(s) to the set $S$?

Options:

(A) $f(x) = \frac{x^2}{2} e^{-x} + (e - \frac{1}{2}) e^{-x}$

(B) $f(x) = -\frac{x^2}{2} e^{-x} + (e + \frac{1}{2}) e^{-x}$

(C) $f(x) = \frac{e^x}{2} (x - \frac{1}{2}) + (e - \frac{e^2}{4}) e^{-x}$

(D) $f(x) = \frac{e^x}{2} (\frac{1}{2} - x) + (e + \frac{e^2}{4}) e^{-x}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a differential equation; we are uncovering the hidden structure of a family of functions.

The problem asks us to explore the set $S$, which contains solutions to the differential equation:

$$\frac{dy}{dx} + \alpha y = x e^{\beta x}$$

This is subject to the initial condition $y(1) = 1$. This is a beautiful journey through the mechanics of linear ODEs.

### Identifying the Beast

First, let us look at the equation: $\frac{dy}{dx} + \alpha y = x e^{\beta x}$. This is a classic First-Order Linear Differential Equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.

By comparing, we immediately see that $P(x) = \alpha$ and $Q(x) = x e^{\beta x}$. This identification is our compass. Without it, we are lost; with it, we have a clear path forward.

### The Magic Key (Integrating Factor)

To solve this, we need an Integrating Factor (IF). The formula is $	ext{IF} = e^{\int P(x) dx}$.

Since $P(x) = \alpha$, our IF is simply $e^{\int \alpha dx} = e^{\alpha x}$. This factor is the 'magic key' that transforms our equation into a form where the left side becomes the derivative of a product.

Multiplying both sides by $e^{\alpha x}$, we get:

$$y e^{\alpha x} = \int (x e^{\beta x} \cdot e^{\alpha x}) dx + C$$

Using the laws of exponents, we combine the terms: $e^{\beta x} \cdot e^{\alpha x} = e^{(\alpha + \beta)x}$. Our equation now rests on the integral $\int x e^{(\alpha + \beta)x} dx$.

### The Fork in the Road

Here is where the intuition of a JEE aspirant shines. The integral $\int x e^{(\alpha + \beta)x} dx$ changes its nature based on the exponent coefficient $k = \alpha + \beta$.

**Case 1: $\alpha + \beta = 0$**

If $\alpha + \beta = 0$, the exponential term $e^{0x}$ becomes $1$. The integral simplifies to $\int x dx = \frac{x^2}{2} + C$.

Our general solution becomes $y e^{\alpha x} = \frac{x^2}{2} + C$. Applying the initial condition $y(1) = 1$, we find:

$$1 \cdot e^{\alpha} = \frac{1}{2} + C \Rightarrow C = e^{\alpha} - \frac{1}{2}$$

Substituting this back, we get $y = \frac{x^2}{2} e^{-\alpha x} + (e^{\alpha} - \frac{1}{2}) e^{-\alpha x}$. If we set $\alpha = 1$, this matches Option A perfectly!

**Case 2: $\alpha + \beta 
eq 0$**

If $\alpha + \beta 
eq 0$, we must use Integration by Parts. Let $u = x$ and $dv = e^{(\alpha + \beta)x} dx$. Then $du = dx$ and $v = \frac{e^{(\alpha + \beta)x}}{\alpha + \beta}$.

The formula $\int u dv = uv - \int v du$ gives us:

$$\int x e^{(\alpha + \beta)x} dx = \frac{x e^{(\alpha + \beta)x}}{\alpha + \beta} - \int \frac{e^{(\alpha + \beta)x}}{\alpha + \beta} dx = \frac{x e^{(\alpha + \beta)x}}{\alpha + \beta} - \frac{e^{(\alpha + \beta)x}}{(\alpha + \beta)^2}$$

Substituting this into our general solution, we get:

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

By testing $\alpha = 1$ and $\beta = 1$, we find that this matches Option C!

### Conclusion

We have navigated the two cases, verified our constants, and matched our results to the options. The beauty of this problem lies in recognizing that the solution is not a single function, but a family of functions defined by the parameters $\alpha$ and $\beta$.

You have successfully decoded the set $S$. Keep this analytical mindset, and no differential equation will ever stand in your way!

Select Answer:

Visualized Solution

Identify the Differential Equation

Calculate Integrating Factor (IF)

Setup the General Solution

Analyze the Exponent

Case 1: $α + β = 0$

Apply Initial Condition $y (1) = 1$

Final Equation for Case 1

Case 2: $α + β \neq = 0$

Apply Integration by Parts

Evaluate the Final Integral

Analyze Option C

Verify Initial Condition for Option C

Final Conclusion

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Identifying the Beast

The Magic Key (Integrating Factor)

The Fork in the Road

Conclusion

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