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MCQMediumJEE 2024Linear Differential Equations

JEE Mathematics 2024 Question with Solution

Let ∫0x1−(y′(t))2 dt=∫0xy(t) dt\int_{0}^{x} \sqrt{1 - (y'(t))^2} \, dt = \int_{0}^{x} y(t) \, dt, 0≤x≤30 \leq x \leq 3, y≥0y \geq 0, y(0)=0y(0) = 0. Then at x=2x = 2, y′′+y+1y'' + y + 1 is equal to:

  • A

    11

  • B

    22

  • C

    2\sqrt{2}

  • D

    12\frac{1}{2}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

∫0x1−(y′(t))2 dt=∫0xy(t) dt\int_{0}^{x} \sqrt{1 - (y'(t))^2} \, dt = \int_{0}^{x} y(t) \, dt

with 0≤x≤30 \leq x \leq 3, y≥0y \geq 0, and y(0)=0y(0)=0.

Find: The value of y′′+y+1y'' + y + 1 at x=2x=2.

Differentiate both sides with respect to xx:

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

Now square both sides:

1−(y′(x))2=y(x)21 - (y'(x))^2 = y(x)^2

Rearranging,

(y′(x))2+y(x)2=1(y'(x))^2 + y(x)^2 = 1

Differentiate again with respect to xx:

2y′(x)y′′(x)+2y(x)y′(x)=02y'(x)y''(x) + 2y(x)y'(x) = 0

So,

y′(x)(y′′(x)+y(x))=0y'(x)(y''(x) + y(x)) = 0

From the solution working, this gives

y′′(x)+y(x)=0y''(x) + y(x) = 0

Therefore,

y′′(2)+y(2)+1=0+1=1y''(2) + y(2) + 1 = 0 + 1 = 1

Hence, the correct option is A.

Detailed Differentiation

Given:

∫0x1−(y′(t))2 dt=∫0xy(t) dt\int_{0}^{x} \sqrt{1 - (y'(t))^2} \, dt = \int_{0}^{x} y(t) \, dt

Find: y′′+y+1y'' + y + 1 at x=2x=2.

By differentiating the integral equation once, we get

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

Since y≥0y \geq 0, squaring is consistent here.

Squaring gives

1−(y′(x))2=y(x)21 - (y'(x))^2 = y(x)^2

which can be written as

(y′(x))2=1−y(x)2(y'(x))^2 = 1 - y(x)^2

Differentiate both sides:

2y′(x)y′′(x)=−2y(x)y′(x)2y'(x)y''(x) = -2y(x)y'(x)

Assuming y′(x)≠0y'(x) \neq 0 in the working shown, divide by 2y′(x)2y'(x):

y′′(x)=−y(x)y''(x) = -y(x)

Now substitute into the required expression:

y′′(2)+y(2)+1=−y(2)+y(2)+1=1y''(2) + y(2) + 1 = -y(2) + y(2) + 1 = 1

Therefore, the value is 11.

Common mistakes

  • Differentiating the integral equation incorrectly. By the Fundamental Theorem of Calculus, the integrands become functions of xx after differentiation. Do not keep the variable as tt in the final differentiated equation.

  • Forgetting to square both sides after obtaining 1−(y′(x))2=y(x)\sqrt{1-(y'(x))^2}=y(x). Without this step, the relation needed to connect y′′y'' and yy cannot be derived. First convert it into an algebraic identity.

  • Stopping at y′(x)(y′′(x)+y(x))=0y'(x)(y''(x)+y(x))=0 and concluding nothing. The provided solution uses the intended branch y′′(x)+y(x)=0y''(x)+y(x)=0. Use this relation to evaluate the required expression.

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