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Integral Equation Reduced to a Linear DE | JEE

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

Question

Let

y = y(x)

satisfy

y(x) + \displaystyle\int_1^x y(t)\,dt = x^2

. The value of

y(e)

is:

2e - 2 + e^{1-e}

correct

2e - 1 + e^{1-e}

2e + 2 + e^{1-e}

2e + 1 + e^{1-e}

Solution

Step 1: Differentiate both sides with respect to

x

y'(x) + y(x) = 2x \qquad (\text{linear}), \quad \text{IF} = e^x.

Step 2: Solve:

y\,e^x = \int 2x\,e^x\,dx = 2(x - 1)e^x + c \Rightarrow y = 2(x - 1) + c\,e^{-x}.

Step 3: From the original relation at

x = 1

y(1) + 0 = 1 \Rightarrow y(1) = 1

. Also

y(1) = 0 + c e^{-1}

, so

c = e

. Hence

y = 2(x - 1) + e^{1-x}.

Step 4: At

x = e

y(e) = 2(e - 1) + e^{1 - e} = 2e - 2 + e^{1-e}.

Correct answer: (1)

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