Differential Equation : $\frac{dy}{dx} +\int_0^5{y\,dx}=27$

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Given $y=f(x)$ , is twice differentiable, passes through the origin and satisfies the equation, $$\frac{dy}{dx} +\int_0^5{y\,dx}=27$$What is the probability that $2$ randomly chosen variables $a$, $b$ from the set $S=\{1,2,3,4\}$ lies on the curve as $(a,b)$?

My Attempt:

As $\int_0^5{y\,dx}$ is a constant.

$$\frac{d^2y}{dx^2}=0$$

Therefore, $y=ax+b$,

As curve passes through the origin, $b=0$, so $y=ax$.

On putting this in the equation, we get $a=2$.

Therefore, $$y=2x.$$

Is my approach right? And also how else can we attempt this question?

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You have $y=2x$ so the only integer combinations from your set $S$ are $(1, 2)$ and $(2, 4)$.

The probability is thus $\dfrac2{4\cdot4}=\dfrac18$.