Solution of the differential equation
`x(dy)/(dx) +2y = x^(2) log x ` is Then
`x^(2)-y^(2)-1+cx=0`. True or False

Answer

To solve the differential equation \( x \frac{dy}{dx} + 2y = x^2 \log x \) and verify if the solution can be expressed as \( x^2 - y^2 - 1 + cx = 0 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ x \frac{dy}{dx} + 2y = x^2 \log x \]
We can divide the entire equation by \( x \) to simplify it:
\[ \frac{dy}{dx} + \frac{2y}{x} = x \log x \]

### Step 2: Identify \( p \) and \( q \)
The equation is now in the standard form of a linear differential equation:
\[ \frac{dy}{dx} + p y = q \]
where \( p = \frac{2}{x} \) and \( q = x \log x \).

### Step 3: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2 \log x} = x^2 \]

### Step 4: Multiply the Equation by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor \( x^2 \):
\[ x^2 \frac{dy}{dx} + 2xy = x^3 \log x \]

### Step 5: Rewrite the Left-Hand Side
The left-hand side can be rewritten as the derivative of a product:
\[ \frac{d}{dx}(x^2 y) = x^3 \log x \]

### Step 6: Integrate Both Sides
Now, we integrate both sides:
\[ \int \frac{d}{dx}(x^2 y) \, dx = \int x^3 \log x \, dx \]
The left-hand side simplifies to:
\[ x^2 y = \int x^3 \log x \, dx \]
To solve the right-hand side, we use integration by parts. Let:
- \( u = \log x \) and \( dv = x^3 dx \)
Then:
- \( du = \frac{1}{x} dx \) and \( v = \frac{x^4}{4} \)

Using integration by parts:
\[ \int x^3 \log x \, dx = \frac{x^4}{4} \log x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 \, dx \]
Calculating the remaining integral:
\[ \int x^3 \, dx = \frac{x^4}{4} \]
Thus:
\[ \int x^3 \log x \, dx = \frac{x^4}{4} \log x - \frac{x^4}{16} + C \]

### Step 7: Solve for \( y \)
Now substituting back, we have:
\[ x^2 y = \frac{x^4}{4} \log x - \frac{x^4}{16} + C \]
Dividing by \( x^2 \):
\[ y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2} \]

### Step 8: Compare with the Given Solution
The given solution is:
\[ x^2 - y^2 - 1 + cx = 0 \]
We can see that the derived solution does not match the form provided in the statement. Therefore, the statement is **False**.

### Conclusion
The solution of the differential equation \( x \frac{dy}{dx} + 2y = x^2 \log x \) does not yield the expression \( x^2 - y^2 - 1 + cx = 0 \).