Question

If \( y(x) \) satisfies the differential equation \( x \frac{dy}{dx} + (x - y) = 0 \) subject to the condition \( y(1) = 0 \), then \( y(e) \) is ...........

• A. $-e$
• B. $-2e$
• C. $e$
• D. $2e$

Hint:

When solving linear differential equations, separating variables can often simplify the integration process. Don't forget to use the initial condition to find the constant of integration.

Answer 1

The Correct Option is A

We are given the differential equation: \[ x \frac{dy}{dx} + (x - y) = 0 \] Rearranging the equation: \[ x \frac{dy}{dx} = y - x \] \[ \frac{dy}{dx} = \frac{y - x}{x} \] This is a first-order linear differential equation. To solve it, we use the method of separation of variables. Rearranging terms: \[ \frac{dy}{y - x} = \frac{dx}{x} \] Now, integrate both sides: \[ \int \frac{dy}{y - x} = \int \frac{dx}{x} \] The integral of $\frac{1}{y - x}$ is $\ln|y - x|$ and the integral of $\frac{1}{x}$ is $\ln|x|$. So we have: \[ \ln|y - x| = \ln|x| + C \] Exponentiate both sides: \[ |y - x| = A|x| \] Where $A = e^C$. Now, solve for $y$: \[ y - x = A x \] \[ y = (A + 1) x \] Using the initial condition $y(1) = 0$: \[ 0 = (A + 1) \times 1 \] \[ A + 1 = 0 ⇒ A = -1 \] Thus, the solution is: \[ y = (-1 + 1)x = 0 \] Substituting $x = e$ into the equation: \[ y(e) = -e \] Thus, the value of $y(e)$ is $-e$.