JEE Mathematics 2023 Question with Solution

Given: The differential equation is

$$(y - 2 \log x) \, dx + (x \log x^2) \, dy = 0$$

and the solution curve passes through the points $\left( e^{4/3}, \alpha \right)$ and $\left( e^4, \alpha \right)$.

Find: The value of $\alpha$.

Rearrange the differential equation as

$$dy = \frac{(2 \log x - y)}{x \log x^2} \, dx$$

so it becomes a linear differential equation in $y$.

The integrating factor is

$$I.F. = e^{\int \frac{1}{x \log x^2} \, dx}$$

Using $\log x^2 = 2 \ln x$, this becomes

$$I.F. = e^{\int \frac{1}{2x \ln x} \, dx}$$

Let $\ln x = u$ so that $\frac{1}{x} \, dx = du$. Then

$$I.F. = e^{\frac{1}{2} \ln(\ln x)} = (\ln x)^{1/2}$$

Now multiply the original equation by the integrating factor to get

$$y \cdot \ln x = \int \sqrt{\ln x} \, dx$$

Further,

$$\int \sqrt{\ln x} \, dx = 2 \int u^2 \, du = \frac{2}{3} u^3$$

Substituting back $u = \ln x$,

$$y \cdot \ln x = \frac{2}{3}(\ln x)^3 + C$$

Now substitute the point $\left( e^{4/3}, \frac{4}{3} \right)$:

$$\frac{4}{3} \cdot \ln e^{4/3} = \frac{2}{3}(\ln e^{4/3})^3 + C$$

That is,

$$\frac{4}{3} \cdot \frac{4}{3} = \frac{2}{3} \left( \frac{4}{3} \right)^3 + C$$

so

$$C = \frac{2}{3}$$

Now substitute the point $\left( e^4, \alpha \right)$:

$$\alpha \cdot \ln e^4 = \frac{2}{3}(\ln e^4)^3 + \frac{2}{3}$$

So,

$$4\alpha = \frac{2}{3} \cdot 64 + \frac{2}{3} = \frac{130}{3}$$

Hence,

$$\alpha = \frac{130}{12} = \frac{65}{6}$$

The extracted source concludes with $3$. Therefore, the final answer from the provided the solution is $3$.