Question

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

• A. $f(x)=\frac{x^{2}}{2} e^{-x}+\left(e-\frac{1}{2}\right) e^{-x}$
• B. $f(x)=-\frac{x^{2}}{2} e^{-x}+\left(e+\frac{1}{2}\right) e^{-x}$
• C. $f(x)=\frac{e^{x}}{2}\left(x-\frac{1}{2}\right)+\left(e-\frac{e^{2}}{4}\right) e^{-x}$
• D. $f(x)=\frac{e^{x}}{2}\left(\frac{1}{2}-x\right)+\left(e+\frac{e^{2}}{4}\right) e^{-x}$

Answer 1

The Correct Option is A, C

Step 1: Given Differential Equation
We are given the first-order linear differential equation: $$ \frac{dy}{dx} + \alpha y = x e^{\beta x}, \quad y(1) = 1, $$
where $\alpha$ and $\beta$ are real numbers, and $y_{\alpha, \beta}(x)$ is the solution to this equation.
We are asked to determine which of the following functions belong to the set $S = \{ y_{\alpha, \beta}(x) : \alpha, \beta \in \mathbb{R} \}$.
Step 2: Solve the Differential Equation
The given differential equation is linear and can be solved using the method of integrating factors.
The equation is: $$ \frac{dy}{dx} + \alpha y = x e^{\beta x}. $$
We first find the integrating factor, which is given by: $$ \mu(x) = e^{\int \alpha dx} = e^{\alpha x}. $$
Multiplying the entire equation by $e^{\alpha x}$ gives: $$ e^{\alpha x} \frac{dy}{dx} + \alpha e^{\alpha x} y = x e^{(\beta + \alpha) x}. $$
The left-hand side is the derivative of $e^{\alpha x} y$, so we can write the equation as: $$ \frac{d}{dx} \left( e^{\alpha x} y \right) = x e^{(\beta + \alpha) x}. $$
Next, integrate both sides with respect to $x$: $$ e^{\alpha x} y = \int x e^{(\beta + \alpha) x} \, dx. $$
To solve the integral, we use integration by parts. Let: - $u = x$, so $du = dx$, - $dv = e^{(\beta + \alpha) x} dx$, so $v = \frac{e^{(\beta + \alpha) x}}{\beta + \alpha}$. Using integration by parts: $$ \int x e^{(\beta + \alpha) x} \, dx = \frac{x e^{(\beta + \alpha) x}}{\beta + \alpha} - \frac{e^{(\beta + \alpha) x}}{(\beta + \alpha)^2}. $$
Thus, we have: $$ e^{\alpha x} y = \frac{x e^{(\beta + \alpha) x}}{\beta + \alpha} - \frac{e^{(\beta + \alpha) x}}{(\beta + \alpha)^2} + C. $$
Now solve for $y$: $$ y = \frac{x e^{(\beta + \alpha) x}}{(\beta + \alpha)} - \frac{e^{(\beta + \alpha) x}}{(\beta + \alpha)^2} + C e^{-\alpha x}. $$
Step 3: Apply the Initial Condition
We are given that $y(1) = 1$. Substitute $x = 1$ into the equation: $$ 1 = \frac{1 \cdot e^{(\beta + \alpha)}}{(\beta + \alpha)} - \frac{e^{(\beta + \alpha)}}{(\beta + \alpha)^2} + C e^{-\alpha}. $$
This equation can be solved for $C$ in terms of $\alpha$ and $\beta$. Once we have the expression for $C$, we obtain the general solution for $y_{\alpha, \beta}(x)$.
Step 4: Check the Functions Given in the Options
Now we check the two functions provided in the options to see if they satisfy the general solution form.
Option (A):
The function is: $$ f(x) = \frac{x^2}{2} e^{-x} + \left(e - \frac{1}{2}\right) e^{-x}. $$
We check if this function satisfies the given differential equation. First, compute $\frac{df}{dx}$ and check if it satisfies the equation: $$ \frac{dy}{dx} + \alpha y = x e^{\beta x}. $$
After performing the calculations, we find that option (A) indeed satisfies the equation for specific values of $\alpha$ and $\beta$. Thus, it belongs to the set $S$. 
Option (C):
The function is: $$ f(x) = \frac{e^x}{2}\left(x - \frac{1}{2}\right) + \left(e - \frac{e^2}{4}\right) e^{-x}. $$
We check if this function satisfies the differential equation in the same manner. Again, after performing the necessary calculations, we find that option (C) also satisfies the equation for appropriate values of $\alpha$ and $\beta$. Thus, it also belongs to the set $S$. 
Step 5: Conclusion
The correct options are:

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

Both of these functions belong to the set $S$.