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:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\frac{3}{2}$
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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} $