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JEE 2023 Mathematics PYQ — Let be the solution of the differential equation with . Then, is … | Mathem Solvex | Mathem Solvex

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $x = x(y)$ be the solution of the differential equation $2(y+2) \log_e (y+2) dx + (x + 4 - 2 \log_e (y+2)) dy = 0, y > -1$ with $x (e^4 - 2) = 1$ . Then, $x (e^9 - 2)$ is equal to:

Choose the correct answer:

A.
$\frac{4}{9}$
B.
$\frac{32}{9}$
(Correct Answer)
C.
$\frac{10}{3}$

Correct Answer:

$\frac{32}{9}$

Explanation

Solution:

Ise linear differential equation $\frac{dx}{dy} + Px = Q$ ke roop mein likhte hain:

\frac{dx}{dy} + \frac{x}{2(y+2) \log_e (y+2)} = \frac{2 \log_e (y+2) - 4}{2(y+2) \log_e (y+2)}

Integrating Factor $(\text{I.F.}) = e^{\int \frac{dy}{2(y+2) \log(y+2)}} = \sqrt{\log(y+2)}$ .

Puri equation solve karne aur boundary condition $x(e^4-2)=1$ lagane par humein $x(y)$ milta hai.

$y = e^9 - 2$ rakhne par, $x = \frac{32}{9}$ aata hai.

Sahi Option: (2)

Explanation

Solution:

Ise linear differential equation $\frac{dx}{dy} + Px = Q$ ke roop mein likhte hain:

\frac{dx}{dy} + \frac{x}{2(y+2) \log_e (y+2)} = \frac{2 \log_e (y+2) - 4}{2(y+2) \log_e (y+2)}

Integrating Factor $(\text{I.F.}) = e^{\int \frac{dy}{2(y+2) \log(y+2)}} = \sqrt{\log(y+2)}$ .

Puri equation solve karne aur boundary condition $x(e^4-2)=1$ lagane par humein $x(y)$ milta hai.

$y = e^9 - 2$ rakhne par, $x = \frac{32}{9}$ aata hai.

Sahi Option: (2)

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