Find the particular solutions of the following problems :
`x^(2)dy-(x^(2)+xy+y^(2))dx=0, y(1)=1`

Answer

To find the particular solution of the differential equation \( x^2 dy - (x^2 + xy + y^2) dx = 0 \) with the initial condition \( y(1) = 1 \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the given differential equation in a more standard form:
\[ x^2 dy = (x^2 + xy + y^2) dx \]
Dividing both sides by \( x^2 \) gives:
\[ dy = \left(1 + \frac{y}{x} + \frac{y^2}{x^2}\right) dx \]

### Step 2: Identify Homogeneity
To check if the equation is homogeneous, we substitute \( x = \lambda x \) and \( y = \lambda y \):
\[ dy = \left(1 + \frac{\lambda y}{\lambda x} + \frac{(\lambda y)^2}{(\lambda x)^2}\right) dx \]
This simplifies to:
\[ dy = \left(1 + \frac{y}{x} + \frac{y^2}{x^2}\right) dx \]
Since the form remains unchanged, the equation is homogeneous.

### Step 3: Substitute \( y = tx \)
Let \( t = \frac{y}{x} \) or \( y = tx \). Then, differentiating gives:
\[ dy = t dx + x dt \]
Substituting \( y \) into the equation gives:
\[ t dx + x dt = \left(1 + t + t^2\right) dx \]
Rearranging leads to:
\[ x dt = \left(1 + t + t^2 - t\right) dx \]
Thus:
\[ x dt = (1 + t^2) dx \]

### Step 4: Separate Variables
Now we separate the variables:
\[ \frac{dt}{1 + t^2} = \frac{dx}{x} \]

### Step 5: Integrate Both Sides
Integrating both sides:
\[ \int \frac{dt}{1 + t^2} = \int \frac{dx}{x} \]
This results in:
\[ \tan^{-1}(t) = \ln|x| + C \]

### Step 6: Substitute Back for \( y \)
Substituting back \( t = \frac{y}{x} \):
\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + C \]

### Step 7: Apply Initial Condition
Using the initial condition \( y(1) = 1 \):
\[ \tan^{-1}(1) = \ln(1) + C \]
\[ \frac{\pi}{4} = 0 + C \implies C = \frac{\pi}{4} \]

### Step 8: Write the Particular Solution
Thus, the particular solution is:
\[ \tan^{-1}\left(\frac{y}{x}\right) = \ln|x| + \frac{\pi}{4} \]

### Step 9: Final Rearrangement
To express \( y \) in terms of \( x \):
\[ \frac{y}{x} = \tan\left(\ln|x| + \frac{\pi}{4}\right) \]
Therefore, the final form of the particular solution is:
\[ y = x \tan\left(\ln|x| + \frac{\pi}{4}\right) \]