Question

If \( \frac{dy}{dx} + y = x \), and \( y(0) = 0 \), then the value of \( y(1) \) is:

• A. \( e \)
• B. \( \frac{1}{e} \)
• C. 1
• D. -1

Hint:

For linear first-order differential equations, use the integrating factor method. This method involves multiplying the equation by an integrating factor to make the left-hand side an exact derivative, allowing you to integrate both sides easily.

Answer 1

The Correct Option is B

The given differential equation is: \[ \frac{dy}{dx} + y = x \] This is a first-order linear differential equation. To solve this, we will use the integrating factor method. The general form of a first-order linear differential equation is: \[ \frac{dy}{dx} + P(x)y = Q(x) \] Here, \( P(x) = 1 \) and \( Q(x) = x \). The integrating factor \( \mu(x) \) is given by: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int 1 \, dx} = e^x \] Now, multiply both sides of the original equation by the integrating factor \( e^x \): \[ e^x \frac{dy}{dx} + e^x y = x e^x \] This simplifies to: \[ \frac{d}{dx} \left( e^x y \right) = x e^x \] Next, integrate both sides with respect to \( x \): \[ \int \frac{d}{dx} \left( e^x y \right) \, dx = \int x e^x \, dx \] The left side is simply \( e^x y \), and for the right side, we use integration by parts: \[ \int x e^x \, dx = e^x (x - 1) + C \] Thus, we have: \[ e^x y = e^x (x - 1) + C \] Now, solve for \( y \): \[ y = x - 1 + C e^{-x} \] We are given the initial condition \( y(0) = 0 \). Substituting \( x = 0 \) and \( y(0) = 0 \) into the equation: \[ 0 = 0 - 1 + C e^{0} \] \[ C - 1 = 0 \quad \Rightarrow \quad C = 1 \] Thus, the solution to the differential equation is: \[ y = x - 1 + e^{-x} \] Finally, we need to find \( y(1) \): \[ y(1) = 1 - 1 + e^{-1} = \frac{1}{e} \] Thus, the value of \( y(1) \) is \( \boxed{\frac{1}{e}} \).