General solution of differential equation $x\frac{dy}{dx}+y~log~y=xye^{x}$ is equal to :

Question

Accepted Answer

To solve the differential equation $ x\frac{dy}{dx} + y \log y = xye^x $, we need to find a general solution that matches the form given in Option 2: $ x \log y = e^x(x-1) + c $.

### Explanation for Option 2:

1. **Rearranging the Equation:**
   The given differential equation is:
   $$
   x\frac{dy}{dx} + y \log y = xye^x
   $$
   This can be rearranged to:
   $$
   x\frac{dy}{dx} = xye^x - y \log y
   $$

2. **Separation of Variables:**
   To solve this, we can attempt to separate variables or use an integrating factor. Notice that the equation can be rewritten as:
   $$
   \frac{dy}{dx} = ye^x - \frac{y \log y}{x}
   $$
   This suggests a substitution might simplify the equation. Let $ v = \log y $, then $ y = e^v $ and $ \frac{dy}{dx} = e^v \frac{dv}{dx} $.

3. **Substitution:**
   Substituting $ y = e^v $ into the equation gives:
   $$
   x e^v \frac{dv}{dx} + e^v v = xe^v e^x
   $$
   Dividing through by $ e^v $ simplifies to:
   $$
   x \frac{dv}{dx} + v = xe^x
   $$

4. **Solving the Linear Differential Equation:**
   This is a linear first-order differential equation in $ v $. The standard form is:
   $$
   \frac{dv}{dx} + \frac{v}{x} = e^x
   $$
   The integrating factor is $ \mu(x) = e^{\int \frac{1}{x} dx} = x $.

5. **Applying the Integrating Factor:**
   Multiply through by the integrating factor:
   $$
   x \frac{dv}{dx} + v = xe^x
   $$
   This can be rewritten as:
   $$
   \frac{d}{dx}(xv) = xe^x
   $$

6. **Integrating Both Sides:**
   Integrate both sides with respect to $ x $:
   $$
   xv = \int xe^x \, dx
   $$
   Using integration by parts, let $ u = x $ and $ dv = e^x dx $, then $ du = dx $ and $ v = e^x $. The integral becomes:
   $$
   \int xe^x \, dx = xe^x - \int e^x \, dx = xe^x - e^x + C
   $$
   Thus:
   $$
   xv = xe^x - e^x + C
   $$

7. **Solution in Terms of $ y $:**
   Substitute back $ v = \log y $:
   $$
   x \log y = e^x(x - 1) + C
   $$

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

### Explanation for Why Other Options are Incorrect:

- **Option 1: $ \log y = xe^x + c $**
  - This option does not account for the $ (x-1) $ term that arises from the integration by parts process.

- **Option 3: $ \frac{1}{x} \log y = e^x(x+1) + c $**
  - This option incorrectly suggests a division by $ x $ and an incorrect term $ (x+1) $ instead of $ (x-1) $.

- **Option 4: $ \log y = e^x(x-1) + c $**
  - This option lacks the multiplication by $ x $ on the left side, which is necessary as derived from the integration process.

Thus, Option 2 is the correct solution to the differential equation, as it accurately reflects the integration and substitution steps required to solve the equation.