Question

If \(x y' + y - e^x = 0, \, y(a) = b\), then
\[ \lim_{x \to 1} y(x) \text{ is} \]

• A. \(e + 2ab - e^a\)
• B. \(e^2 + ab - e^{-a}\)
• C. \(e - ab + e^a\)
• D. \(e + ab - e^a, \, y' = \frac{dy}{dx}\)

Hint:

For first-order linear differential equations, use an integrating factor to simplify and solve the equation.

Answer 1

The Correct Option is D

Step 1: Start with the given differential equation:

\[ x y' + y - e^x = 0 \]

Rearranging it:

\[ x y' = e^x - y \]

This is a linear first-order differential equation.

Step 2: Multiply the equation by the integrating factor \( \mu(x) \), which is \( \mu(x) = x \).

\[ x \cdot y' + x \cdot y = x \cdot e^x \]

This simplifies to:

\[ \frac{d}{dx}(xy) = x e^x \]

Step 3: Integrate both sides of the equation with respect to \( x \):

\[ \int \frac{d}{dx}(xy) dx = \int x e^x dx \]

The right-hand side can be integrated by parts:

\[ \int x e^x dx = x e^x - e^x + C \]

Thus, we have:

\[ xy = x e^x - e^x + C \]

Step 4: Solve for \( y \):

\[ y = e^x - \frac{e^x}{x} + \frac{C}{x} \]

Step 5: Use the initial condition \( y(a) = b \) to find \( C \):

\[ b = e^a - \frac{e^a}{a} + \frac{C}{a} \]

Solving for \( C \):

\[ C = a \left( b - e^a + \frac{e^a}{a} \right) \]

Step 6: Substitute the value of \( C \) into the equation for \( y \):

\[ y = e^x - \frac{e^x}{x} + \frac{a \left( b - e^a + \frac{e^a}{a} \right)}{x} \]

Step 7: Finally, take the limit as \( x \to 1 \):

\[ \lim_{x \to 1} y(x) = e + ab - e^a \]

Answer: (D) \( e + ab - e^a \)

Answer 2

Solving $xy' + y - e^x = 0$ and Finding $\lim_{x \to 1} y(x)$

We are given the first-order linear differential equation:

$$ xy' + y - e^x = 0 $$

with the initial condition $y(a) = b$. We need to find $\lim_{x \to 1} y(x)$ based on the hint provided.

Step 1: Rewrite the differential equation

$$ xy' + y = e^x $$

Step 2: Recognize the left side as the derivative of a product

The left side is the derivative of $(xy)$ with respect to $x$:

$$ \frac{d}{dx}(xy) = xy' + y $$

Step 3: Integrate both sides with respect to $x$

$$ \int \frac{d}{dx}(xy) dx = \int e^x dx $$

$$ xy = e^x + C $$

where $C$ is the constant of integration.

Step 4: Use the initial condition $y(a) = b$

$$ a \cdot y(a) = e^a + C $$

$$ ab = e^a + C $$

$$ C = ab - e^a $$

Step 5: Substitute the value of $C$ back into the general solution

$$ xy = e^x + ab - e^a $$

Step 6: Solve for $y(x)$

$$ y(x) = \frac{e^x + ab - e^a}{x} $$

Step 7: Find the limit as $x \to 1$ (based on the hint)

$$ \lim_{x \to 1} y(x) = \lim_{x \to 1} \frac{e^x + ab - e^a}{x} $$

Substitute $x = 1$ into the expression for $y(x)$:

$$ \lim_{x \to 1} y(x) = \frac{e^1 + ab - e^a}{1} = e + ab - e^a $$

Based on the hint's final limit calculation, the value of the limit is $e + ab - e^a$.

Final Answer: (D) $e + ab - e^a$, $\left( y' = \frac{dy}{dx} \right)$