If $x\frac{dy}{dx}+y~log~y=xye^{x}$, then

Question

Accepted Answer

To solve the given differential equation $ x\frac{dy}{dx} + y \log y = xye^{x} $, we need to verify why Option 2 is the correct solution and why the other options are incorrect.

### Explanation for Option 2:

The given differential equation is:

$$ x\frac{dy}{dx} + y \log y = xye^{x} $$

This can be rewritten as:

$$ \frac{dy}{dx} + \frac{y \log y}{x} = ye^{x} $$

This is a first-order linear differential equation in the standard form:

$$ \frac{dy}{dx} + P(x)y = Q(x) $$

where $ P(x) = \frac{\log y}{x} $ and $ Q(x) = ye^{x} $.

To solve this, we use an integrating factor. However, the presence of $ \log y $ complicates the direct application of standard methods. Instead, we can try a substitution method. Let $ v = \log y $, then $ y = e^{v} $ and $ \frac{dy}{dx} = e^{v} \frac{dv}{dx} $.

Substituting these into the equation, we get:

$$ x e^{v} \frac{dv}{dx} + e^{v} v = x e^{v} e^{x} $$

Dividing throughout by $ e^{v} $, we have:

$$ x \frac{dv}{dx} + v = x e^{x} $$

This simplifies to:

$$ \frac{dv}{dx} + \frac{v}{x} = e^{x} $$

This is now a linear differential equation in $ v $. The integrating factor is:

$$ \mu(x) = e^{\int \frac{1}{x} dx} = e^{\log x} = x $$

Multiplying through by the integrating factor:

$$ x \frac{dv}{dx} + v = x e^{x} $$

This can be written as:

$$ \frac{d}{dx}(xv) = x e^{x} $$

Integrating both sides with respect to $ x $:

$$ xv = \int x e^{x} dx $$

Using integration by parts for the right-hand side:

Let $ u = x $ and $ dv = e^{x} dx $, then $ du = dx $ and $ v = e^{x} $.

$$ \int x e^{x} dx = x e^{x} - \int e^{x} dx = x e^{x} - e^{x} + C $$

Thus:

$$ xv = x e^{x} - e^{x} + C $$

Substituting back $ v = \log y $:

$$ x \log y = x e^{x} - e^{x} + C $$

Rearranging gives:

$$ x \log y = e^{x}(x - 1) + C $$

This matches Option 2: $ x \log y = e^{x}(x - 1) + c $.

### Explanation for Incorrect Options:

- **Option 1: $ \log y = xe^{x} + c $**

This option does not account for the $ -e^{x} $ term that arises from the integration by parts process. Therefore, it does not satisfy the differential equation.

- **Option 3: $ x^{-1} \log y = e^{x}(x + 1) + c $**

This option incorrectly suggests a division by $ x $ on the left-hand side, which is not consistent with the integration process and the structure of the differential equation.

- **Option 4: $ x \log y = e^{x}(x + 2) + c $**

This option incorrectly adds an extra term $ +2 $ in the expression, which does not appear in the integration process or the solution derived from the differential equation.

In conclusion, Option 2 is the correct solution as it accurately represents the integration and solution process for the given differential equation.