JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $y = y(x)$ be the solution curve of $\frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \log_e x))$ with $y(1) = 3$ . Then $\frac{y^2(x)}{9}$ is equal to:

Choose the correct answer:

Explanation

Differential Equation Solution

Given,

\frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \ln x)), y(1) = 3

\Rightarrow \frac{dy}{dx} = \frac{y}{x} + y^3(1 + \ln x)

\Rightarrow \frac{1}{y^3} \frac{dy}{dx} - \frac{1}{y^2} \left( \frac{1}{x} \right) = 1 + \ln x

Let $-\frac{1}{y^2} = u \Rightarrow \frac{1}{y^3} \frac{dy}{dx} = \frac{1}{2} \frac{du}{dx}$

\Rightarrow \frac{1}{2} \frac{du}{dx} + \frac{u}{x} = 1 + \ln x

\Rightarrow \frac{du}{dx} + \frac{2u}{x} = 2(1 + \ln x), \text{ which is a linear differential equation.}

Now, $\text{I.F.} = e^{\int \frac{2}{x} dx}$

\Rightarrow \text{I.F.} = e^{2 \ln x} = x^2

So, the solution of the given differential equation is:

u (\text{I.F.}) = \int 2(1 + \ln x)(\text{I.F.}) dx + C

\Rightarrow ux^2 = \int 2x^2(1 + \ln x) dx + C

\Rightarrow ux^2 = 2(1 + \ln x) \frac{x^3}{3} - 2 \int \left( \frac{1}{x} \right) \cdot \frac{x^3}{3} dx + C

\Rightarrow ux^2 = \frac{2x^3}{3}(1 + \ln x) - \frac{2x^3}{9} + C

\Rightarrow -\frac{1}{y^2} x^2 = \frac{2x^3}{3}(1 + \ln x) - \frac{2x^3}{9} + C

Put $x = 1$ and $y = 3$ in the above equation, we get:

-\frac{1}{9}(1) = \frac{2}{3}(1 + 0) - \frac{2}{9} + C

-\frac{1}{9}(1) = \frac{2}{3}(1 + 0) - \frac{2}{9} + C

Choose the correct answer:

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