JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{5}{x(x^5 + 1)}y = \frac{(x^5 + 1)^2}{x^7}, x > 0$ . If $y(1) = 2$ , then $y(2)$ is equal to:

Choose the correct answer:

Explanation

Solution

Yeh ek Linear Differential Equation hai: $\frac{dy}{dx} + P(x)y = Q(x)$ .

Step 1: Integrating Factor ( $I.F.$ ) nikalna

P(x) = \frac{5}{x(x^5 + 1)} = \frac{5x^4}{x^5(x^5 + 1)}

I.F. = e^{\int \frac{5x^4}{x^5(x^5 + 1)} dx}

Maan lijiye $x^5 = t \implies 5x^4 dx = dt$ :

I.F. = e^{\int \frac{dt}{t(t+1)}} = e^{\ln|\frac{t}{t+1}|} = \frac{x^5}{x^5 + 1}

Step 2: General Solution

y \cdot \left(\frac{x^5}{x^5 + 1}\right) = \int \frac{(x^5 + 1)^2}{x^7} \cdot \frac{x^5}{x^5 + 1} dx

y \cdot \frac{x^5}{x^5 + 1} = \int \frac{x^5 + 1}{x^2} dx = \int (x^3 + x^{-2}) dx

y \cdot \frac{x^5}{x^5 + 1} = \frac{x^4}{4} - \frac{1}{x} + C

Step 3: Constant $C$ aur $y(2)$ nikalna

$y(1) = 2$ rakhne par: $2 \cdot \frac{1}{2} = \frac{1}{4} - 1 + C \implies C = \frac{7}{4}$ .

Ab $x = 2$ rakhne par:

y(2) \cdot \frac{32}{33} = \frac{16}{4} - \frac{1}{2} + \frac{7}{4} = \frac{21}{4}

y(2) = \frac{21}{4} \cdot \frac{33}{32} = \frac{693}{128}

Sahi Option: (1)

Choose the correct answer:

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