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Integro-Differential Equation and Greatest Integer | JEE

JEE Maths question with a full step-by-step solution.

Question

If the solution of

\dfrac{dy}{dx} = y + \displaystyle\int_0^2 y\,dx

y(x)

with

y(0) = 1

, then

\big[\,|y(2)|\,\big]

equals (where

[\,\cdot\,]

is the greatest integer function):

2

3

4

correct

5

Solution

Step 1:

\displaystyle\int_0^2 y\,dx = K

is a constant, so

\dfrac{dy}{dx} = y + K

. Separate and integrate:

\frac{dy}{y + K} = dx \Rightarrow \ln(y + K) = x + \lambda.

Using

y(0) = 1

\lambda = \ln(1 + K)

, hence

y + K = (1 + K)e^x

, i.e.

y = (1+K)e^x - K

. Step 2: Determine

K

K = \int_0^2\big[(1+K)e^x - K\big]dx = (1+K)(e^2 - 1) - 2K \Rightarrow 3K = (1+K)(e^2 - 1).

\Rightarrow K(4 - e^2) = e^2 - 1 \Rightarrow K = \frac{e^2 - 1}{4 - e^2}.

Step 3: Evaluate

y(2)

. Since

1 + K = \dfrac{3}{4 - e^2}

y(2) = (1+K)e^2 - K = \frac{3e^2 - (e^2 - 1)}{4 - e^2} = \frac{2e^2 + 1}{4 - e^2}.

Numerically

|y(2)| = \dfrac{2e^2 + 1}{e^2 - 4} \approx \dfrac{15.78}{3.39} \approx 4.66

, so

\big[\,|y(2)|\,\big] = 4

. Correct answer: (3)

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