JEE Mathematics 2023 Question with Solution

Given:

$$\frac{dy}{dx} = \frac{y}{x}(1 + xy^2(1 + \log x)), \qquad x>0, \qquad y(1)=3$$

Find: $\frac{y^2(x)}{9}$ and identify the correct option.

From the extracted working,

$$\frac{dx}{dy} - \frac{x}{y} = y^3(1+\log x)$$

and then

$$\frac{1}{y^3}\frac{dx}{dy} - \frac{x}{y^2} = 1+\log x$$

Let

$$t = -\frac{1}{y^2}$$

Then

$$\frac{y^3}{2}\frac{dx}{dy} = \frac{dx}{dt}$$

so the equation becomes

$$\frac{dx}{dt} + \frac{x}{2t} = 2(1+\log x)$$

The integrating factor used in the extracted solution is

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

After integration, the working gives

$$\frac{y^2}{-x^2} = \frac{3}{2}\left((1+\log x)x^3 - \frac{3}{x^3}\right) + C$$

Using the initial condition $y(1)=3$, the final result stated in the solution is

$$\frac{y^2}{9} = \frac{x^2}{5 - 2x^3(2 + \log x^3)}$$

Therefore, the correct option is A.

The second provided approach states option A, but its intermediate derivation is inconsistent with the original differential equation. Still, its final conclusion matches the primary the solution.

It concludes directly that

$$\frac{y^2}{9} = \frac{x^2}{5 - 2x^3(2 + \log x^3)}$$

Hence, the correct option is A.

Note: The option text and the extracted source contain some bracket inconsistencies, but the solution explicitly marks A as correct.