Let ∫_0^x √(1-(y'(t))²) dt = ∫_0^x y(t) dt, 0 ≤ x ≤ 3, y ≥ 0, y(0) = 0. Then at x=2, y'' + y + 1 is

Q: Why can we just differentiate the integral? Doesn't that change the equation?

That is a fantastic question! We aren't changing the equation; we are transforming it. Since the equality holds for all $x$ in the given interval, the rates of change (derivatives) of both sides must also be equal. It is like saying if $A=B$, then $\frac{dA}{dx} = \frac{dB}{dx}$. By differentiating, we are simply looking at the 'velocity' of the functions, which allows us to strip away the integral sign.

Q: How do I know to square both sides? Is there a rule for that?

In differential equations, we often need to isolate the derivative to make the equation solvable. When you see a square root trapping your derivative, squaring is a standard algebraic maneuver to 'free' it. Just be mindful of the domain—here, $y \ge 0$ ensures our square root is well-defined and our steps remain valid.

Q: Why did we get $\sin^{-1}(y) = x$? What happened to the constant of integration?

We didn't lose it! We found it using the initial condition $y(0) = 0$. When we integrate $\frac{dy}{\sqrt{1-y^2}} = dx$, we get $\sin^{-1}(y) = x + C$. By plugging in $x=0$ and $y=0$, we get $\sin^{-1}(0) = 0 + C$, which simplifies to $0 = 0 + C$. Thus, $C$ must be zero.

Q: Wait, the question asks for $y'' + y + 1$ at $x=2$. Does the value change if $x$ changes?

That is the beauty of this problem! When you substitute $y = \sin x$ and $y'' = -\sin x$ into the expression, the trigonometric terms cancel out completely. The result is a constant, which means it is the same regardless of the value of $x$! It is a constant function, so $x=2$ is just a distraction.

Question

Let $\int_0^x \sqrt{1-(y'(t))^2} dt = \int_0^x y(t) dt, 0 \le x \le 3, y \ge 0, y(0) = 0$. Then at $x=2, y'' + y + 1$ is equal to

Options:

(A) 1

(B) 2

(C) $\sqrt{2}$

(D) 1/2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a fortress.

You have an integral equation:

$$\int_{0}^{x} \sqrt{1-(y'(t))^2} dt = \int_{0}^{x} y(t) dt$$

The unknown function $y(t)$ is trapped behind integral signs on both sides, and we are given the boundary condition $y(0) = 0$. But here is the secret: in mathematics, every cage has a key. Our key today is the **Newton-Leibniz Rule**.

### The Leibniz Key

To free $y(x)$ from these integral cages, we must perform a surgical operation: differentiation. We differentiate both sides with respect to $x$.

The Newton-Leibniz Rule tells us that the derivative of the integral from zero to $x$ of a function is simply the function evaluated at $x$. Applying $\frac{d}{dx}$ to both sides, the integral signs vanish, leaving us with a much cleaner differential equation:

$$\sqrt{1-(y'(x))^2} = y(x)$$

### The Algebraic Dance

Now, look at that square root. To liberate our derivative $y'$, we square both sides:

$$1 - (y')^2 = y^2$$

Rearranging the terms to isolate the derivative, we obtain $(y')^2 = 1 - y^2$. Taking the square root, we arrive at:

$$y' = \sqrt{1 - y^2}$$

This is a first-order differential equation, and it is begging to be solved using the method of **variable separation**.

### The Path to Sine

We rewrite $y'$ as $\frac{dy}{dx}$ and separate the variables:

$$\frac{dy}{\sqrt{1 - y^2}} = dx$$

Integrating both sides, we recall the classic result from your calculus toolkit:

$$\sin^{-1}(y) = x + C$$

Using our initial condition $y(0) = 0$, we find that $C = 0$. Thus, $\sin^{-1}(y) = x$, which means our mystery function is **$y(x) = \sin x$**.

### The Grand Finale

We have found the function! The question asks for the value of $y'' + y + 1$ at $x=2$.

First, we differentiate $y = \sin x$ twice:

$$y' = \cos x$$
$$y'' = -\sin x$$

Now, substitute these into our target expression:

$$(-\sin x) + (\sin x) + 1$$

Look at that beautiful cancellation! The sine terms vanish, leaving us with the constant $1$.

The value is independent of $x$, so even at $x=2$, the final answer is **$1$**. You have conquered the integral, solved the differential equation, and found the hidden constant. Well done!

Animated Solution for Mathematics - Differential Equations: Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

Select Answer:

Visualized Solution

Analyze the Integral Equation

Introduce Leibniz Rule

Apply Leibniz Rule

Simplify the Equation

Isolate the Derivative

Variable Separable Form

Integrate Both Sides

Find the Constant of Integration

Determine the Function $y (x)$

Calculate Derivatives

Final Evaluation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Leibniz Key

The Algebraic Dance

The Path to Sine

The Grand Finale

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∫0x​1−(y′(t))2​dt=∫0x​y(t)dt,0≤x≤3,y≥0,y(0)=0. Then at x=2,y′′+y+1 is equal to