JEE Mathematics 2023 Question with Solution

Given: $2x^2 y \, dy - (1 - xy^2) \, dx = 0$ with $x > 0$ and $y(2) = \sqrt{\ln 2}$.

Find: $\alpha + \beta - \gamma$ if $\alpha x = \exp(x^\beta y^\gamma)$.

Rearrange the differential equation:

$$2x^2 y \, dy = (1 - xy^2) \, dx$$

So,

$$\frac{dy}{dx} = \frac{1 - xy^2}{2x^2 y}$$

Now substitute

$$t = y^2$$

Then

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

Hence the equation becomes

$$x^2 \frac{dt}{dx} = 1 - xt$$

or

$$\frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^2}$$

This is a linear differential equation in $t$. Its integrating factor is

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

Multiplying throughout by $x$,

$$x \frac{dt}{dx} + t = \frac{1}{x}$$

Thus,

$$\frac{d}{dx}(tx) = \frac{1}{x}$$

Integrating both sides,

$$tx = \int \frac{1}{x} \, dx = \ln x + C$$

So,

$$t = \frac{\ln x + C}{x}$$

Since $t = y^2$,

$$xy^2 = \ln x + C$$

Use the condition $y(2) = \sqrt{\ln 2}$:

$$2(\ln 2) = \ln 2 + C$$

Therefore,

$$C = \ln 2$$

Hence,

$$xy^2 = \ln x + \ln 2 = \ln(2x)$$

So,

$$y^2 = \frac{\ln(2x)}{x}$$

Now evaluate $\exp(x^\beta y^\gamma)$. From the obtained form,

$$x y^2 = \ln(2x)$$

Therefore,

$$\exp(x y^2) = \exp(\ln(2x)) = 2x$$

Comparing with

$$\alpha x = \exp(x^\beta y^\gamma)$$

we get

$$\alpha = 2, \quad \beta = 1, \quad \gamma = 2$$

Thus,

$$\alpha + \beta - \gamma = 2 + 1 - 2 = 1$$

Therefore, the correct option is A.

Given: $2x^2 y \, dy - (1 - xy^2) \, dx = 0$.

Find: the value of $\alpha + \beta - \gamma$.

The key observation is that the terms $y \, dy$ and $xy^2$ suggest the substitution $t = y^2$. Then $dt = 2y \, dy$, which converts the equation into a first-order linear differential equation.

Starting from

$$2x^2 y \, dy - (1 - xy^2) \, dx = 0$$

write it as

$$2x^2 y \, dy = (1 - xy^2) \, dx$$

Divide by $dx$:

$$2x^2 y \frac{dy}{dx} = 1 - xy^2$$

Now since

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

we obtain

$$x^2 \frac{dt}{dx} = 1 - xt$$

which gives

$$\frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^2}$$

Using integrating factor $x$:

$$\frac{d}{dx}(xt) = \frac{1}{x}$$

Integrate:

$$xt = \ln x + C$$

Replace $t$ by $y^2$:

$$xy^2 = \ln x + C$$

Using $y(2) = \sqrt{\ln 2}$,

$$2(\ln 2) = \ln 2 + C$$

so

$$C = \ln 2$$

and therefore

$$xy^2 = \ln(2x)$$

Now match this with the exponential form. Since

$$\exp(xy^2) = \exp(\ln(2x)) = 2x$$

we identify

$$x^\beta y^\gamma = xy^2$$

Hence,

$$\beta = 1, \quad \gamma = 2, \quad \alpha = 2$$

Finally,

$$\alpha + \beta - \gamma = 1$$

So the correct option is A. Note that the solution labels option C, but its own working clearly gives the value $1$, which corresponds to A in the provided options.