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Integro-Differential Equation with Probability | JEE

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Question
Let the curve y=f(x)y = f(x) pass through the origin and satisfy dydx+∫05y dx=27\dfrac{dy}{dx} + \displaystyle\int_0^5 y\,dx = 27. If aa and bb are chosen randomly from S={1,2,3,4}S = \{1, 2, 3, 4\} with replacement, the probability that the curve passes through (a,b)(a, b) is:
A12\dfrac{1}{2}
B16\dfrac{1}{6}
C18\dfrac{1}{8}correct
D112\dfrac{1}{12}
Solution
Step 1: ∫05y dx\displaystyle\int_0^5 y\,dx is a constant; call it KK. Then dydx=27−K\dfrac{dy}{dx} = 27 - K, so y=(27−K)xy = (27 - K)x (the origin gives constant 00). Step 2: Determine KK:
K=∫05(27−K)x dx=(27−K)⋅252⇒2K=25(27−K)⇒27K=675⇒K=25.K = \int_0^5 (27-K)x\,dx = (27-K)\cdot\frac{25}{2} \Rightarrow 2K = 25(27 - K) \Rightarrow 27K = 675 \Rightarrow K = 25.
∴y=(27−25)x=2x.\therefore y = (27 - 25)x = 2x.
 Step 3: The curve y=2xy = 2x passes through (a,b)(a, b) when b=2ab = 2a. Over SS, the favourable pairs are (1,2)(1, 2) and (2,4)(2, 4), so
P=24×4=18.P = \frac{2}{4 \times 4} = \frac{1}{8}.
 Correct answer: (3)
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