سوال ۳۳ کنکور ارشد علوم کامپیوتر ۱۴۰۰

We are given $y$ implicitly as a function of $x$ through

$x = \int_0^y \frac{dt}{\sqrt{a^2 + b^2 t^2}}.$

Step 1 — Differentiate the integral relation. By the Fundamental Theorem of Calculus, the upper limit $y$ is itself a function of $x$, so differentiating both sides with respect to $x$:

$1 = \frac{1}{\sqrt{a^2 + b^2 y^2}} \cdot \frac{dy}{dx}.$

Hence

$y'(x) = \sqrt{a^2 + b^2 y^2}.$

Step 2 — Differentiate again. Apply the chain rule to $y' = \sqrt{a^2 + b^2 y^2}$:

$y''(x) = \frac{1}{2\sqrt{a^2 + b^2 y^2}} \cdot \bigl(2 b^2 y\, y'\bigr) = \frac{b^2 y\, y'}{\sqrt{a^2 + b^2 y^2}}.$

Substituting $y' = \sqrt{a^2 + b^2 y^2}$, the radical cancels:

$y''(x) = \frac{b^2 y \cdot \sqrt{a^2 + b^2 y^2}}{\sqrt{a^2 + b^2 y^2}} = b^2 y.$

Step 3 — Form the ratio.

$\frac{y''(x)}{y(x)} = \frac{b^2 y}{y} = b^2.$

The answer is $b^2$, option 4.

Exam trick: Squaring the first-order relation gives $(y')^2 = a^2 + b^2 y^2$; differentiating $(y')^2$ as $2y'y''$ and dividing by $2y'$ instantly yields $y'' = b^2 y$, which is the constant-coefficient ODE $y'' = b^2 y$ (so $\frac{y''}{y}$ is the constant $b^2$).