Question

If $ \frac{dy}{dx} = y + 3 \quad \text{and} \quad y(0) = 2, \quad \text{then} \quad y(\log 2) \text{ is equal to:} $

• A. 5
• B. 13
• C. -2
• D. 7

Hint:

When solving first-order linear differential equations, always use the method of separation of variables, and remember to apply initial conditions to find the constant of integration.

Answer 1

The Correct Option is D

We are given the differential equation: \[ \frac{dy}{dx} = y + 3 \] This is a first-order linear differential equation. To solve it, we use the method of separation of variables. First, rewrite the equation as: \[ \frac{dy}{dx} = y + 3 \quad \Rightarrow \quad \frac{dy}{y + 3} = dx \] Now integrate both sides: \[ \int \frac{1}{y + 3} \, dy = \int dx \] The left-hand side gives \( \ln |y + 3| \), and the right-hand side gives \( x + C \), where \( C \) is the constant of integration. Thus, the solution is: \[ \ln |y + 3| = x + C \] Exponentiating both sides: \[ |y + 3| = e^{x + C} = e^C \cdot e^x \] Let \( e^C = A \), where \( A \) is a new constant. So, we have: \[ |y + 3| = A e^x \] Since \( y + 3 \) is positive for all values of \( x \), we can drop the absolute value sign: \[ y + 3 = A e^x \] Thus, the general solution is: \[ y = A e^x - 3 \]
Step 1: Apply initial condition \( y(0) = 2 \)
Substitute \( x = 0 \) and \( y(0) = 2 \) into the general solution: \[ 2 = A e^0 - 3 \quad \Rightarrow \quad 2 = A - 3 \] Thus, \( A = 5 \). So, the solution becomes: \[ y = 5 e^x - 3 \]
Step 2: Find \( y(\log 2) \)
Substitute \( x = \log 2 \) into the solution: \[ y(\log 2) = 5 e^{\log 2} - 3 \] Since \( e^{\log 2} = 2 \), we have: \[ y(\log 2) = 5 \times 2 - 3 = 10 - 3 = 7 \] Thus, the correct answer is \( 7 \).