Let $y(x) = ax^2 + bx + 5$ be the solution of the differential equation $\frac{dy}{dx} = x + 1; y(0) = 5$, then the value of \( a + b \) is:

$\frac{3}{2}$

The correct answer is Option (1) → $\frac{3}{2}$

*****

Given $ y(x) = ax^{2} + bx + 5 $

$ \frac{dy}{dx} = 2ax + b $

Differential equation: $ \frac{dy}{dx} = x + 1 $

So, $ 2ax + b = x + 1 $

Equating coefficients: $ 2a = 1 \text{ implies }a = \frac{1}{2}, \quad b = 1 $

Also, $ y(0) = 5 $ is satisfied.

$ a+b = \frac{1}{2} + 1 = \frac{3}{2} $

Answer: $ \frac{3}{2} $