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) \( e + ab - e^a \)