Question

Consider the equation $\dfrac{dy}{dx} + ay = \sin(\omega x)$, where $a$ and $\omega$ are constants. Given $y=1$ at $x=0$, select all correct statements as $x \to \infty$.

• A. $y \to 0$ if $a \neq 0$
• B. $y \to 1$ if $a=0$
• C. $y \to A \exp(|a|x)$ if $a < 0$ ; $A$ is a constant
• D. $y \to B \sin(\omega x + C)$ if $a>0$ ; $B, C$ are constants

Hint:

Always solve first-order linear ODEs using the integrating factor method. Carefully analyze asymptotic behavior separately for $a>0$, $a=0$, and $a<0$.

Answer 1

The Correct Option is B, D

Step 1: Standard form.
\[ \frac{dy}{dx} + ay = \sin(\omega x) \] This is a linear ODE with integrating factor $e^{ax}$.

Step 2: General solution.
\[ y(x) = e^{-ax}\left[ \int e^{ax}\sin(\omega x)dx + C \right] \]

Step 3: Evaluate the integral.
\[ \int e^{ax}\sin(\omega x)dx = \frac{e^{ax}(a\sin(\omega x) - \omega \cos(\omega x))}{a^2 + \omega^2} \] So, \[ y(x) = e^{-ax}\left[ \frac{e^{ax}(a\sin(\omega x) - \omega \cos(\omega x))}{a^2 + \omega^2} + C \right] \] \[ = \frac{a\sin(\omega x) - \omega \cos(\omega x)}{a^2 + \omega^2} + Ce^{-ax} \]

Step 4: Apply initial condition.
At $x=0$, $y(0)=1$: \[ 1 = \frac{-\omega}{a^2 + \omega^2} + C \] \[ C = 1 + \frac{\omega}{a^2 + \omega^2} \] So full solution: \[ y(x) = \frac{a\sin(\omega x) - \omega \cos(\omega x)}{a^2 + \omega^2} + \left(1 + \frac{\omega}{a^2 + \omega^2}\right)e^{-ax} \]

Step 5: Behavior as $x \to \infty$.
- If $a>0$, exponential term $e^{-ax} \to 0$. Remaining solution: \[ y(x) \approx \frac{a}{a^2+\omega^2}\sin(\omega x) - \frac{\omega}{a^2+\omega^2}\cos(\omega x) \] This is of the form $B \sin(\omega x + C)$. So (D) is correct. - If $a=0$, solution reduces to \[ \frac{dy}{dx} = \sin(\omega x), y(0)=1 \] \[ y(x) = 1 - \frac{\cos(\omega x)}{\omega} + \frac{1}{\omega} \] This oscillates around $1$ as $x \to \infty$, so (B) is true. - If $a<0$, exponential term grows unbounded ($e^{|a|x}$), so (C) would be correct. But note: coefficient is fixed; solution diverges, not tends to simple exponential form stated. Thus (C) is not correct. - (A) $y \to 0$ if $a \neq 0$ is false because for $a>0$, $y$ oscillates; for $a<0$, diverges. \[ \boxed{\text{Correct statements: (B) and (D)}} \]