Solve the following differential equation:
`x^(2)dy=(xy+y^(2))dx=0`, when `x=1 and y=1`

Answer

To solve the differential equation given by \( x^2 dy + (xy + y^2) dx = 0 \) with the initial condition \( x = 1 \) and \( y = 1 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ x^2 dy + (xy + y^2) dx = 0 \]
We can rearrange this to isolate \( dy \):
\[ x^2 dy = - (xy + y^2) dx \]
Dividing both sides by \( x^2 \):
\[ dy = -\frac{xy + y^2}{x^2} dx \]

### Step 2: Expressing \( y \) in Terms of \( x \)
Next, we can let \( y = \lambda x \), where \( \lambda \) is a function of \( x \). Then, we differentiate \( y \):
\[ dy = \lambda dx + x \frac{d\lambda}{dx} \]
Substituting this into our rearranged equation gives:
\[ \lambda dx + x \frac{d\lambda}{dx} = -\frac{x(\lambda x) + (\lambda x)^2}{x^2} dx \]
This simplifies to:
\[ \lambda + x \frac{d\lambda}{dx} = -\frac{\lambda + \lambda^2}{x} \]

### Step 3: Simplifying the Equation
Multiplying through by \( x \) to eliminate the fraction:
\[ x\lambda + x^2 \frac{d\lambda}{dx} = -(\lambda + \lambda^2) \]
Rearranging gives:
\[ x^2 \frac{d\lambda}{dx} = -\lambda - \lambda^2 - x\lambda \]
Factoring out \( -\lambda \):
\[ x^2 \frac{d\lambda}{dx} = -\lambda(\lambda + x + 1) \]

### Step 4: Separating Variables
We can separate variables:
\[ \frac{d\lambda}{\lambda(\lambda + x + 1)} = -\frac{dx}{x^2} \]

### Step 5: Integrating Both Sides
Now we integrate both sides. The left side requires partial fraction decomposition:
\[ \frac{1}{\lambda(\lambda + x + 1)} = \frac{A}{\lambda} + \frac{B}{\lambda + x + 1} \]
Solving for \( A \) and \( B \) gives:
\[ 1 = A(\lambda + x + 1) + B\lambda \]
Setting \( \lambda = 0 \) gives \( A = 1/(x + 1) \) and setting \( \lambda = -x - 1 \) gives \( B = -1/(x + 1) \).

Thus, we can integrate:
\[ \int \left( \frac{1}{\lambda} - \frac{1}{\lambda + x + 1} \right) d\lambda = -\int \frac{dx}{x^2} \]
This leads to:
\[ \ln|\lambda| - \ln|\lambda + x + 1| = \frac{1}{x} + C \]

### Step 6: Solving for \( C \)
Now, we substitute back \( \lambda = \frac{y}{x} \):
\[ \ln\left|\frac{y}{x}\right| - \ln\left|\frac{y}{x} + x + 1\right| = -\frac{1}{x} + C \]
Exponentiating both sides gives:
\[ \frac{y}{y + x^2 + x} = Ce^{-\frac{1}{x}} \]

### Step 7: Applying Initial Conditions
Using the initial condition \( x = 1 \) and \( y = 1 \):
\[ \frac{1}{1 + 1 + 1} = Ce^{-1} \]
This gives \( C = \frac{e}{3} \).

### Final Step: Final Equation
Substituting back gives:
\[ \frac{y}{y + x^2 + x} = \frac{e}{3} e^{-\frac{1}{x}} \]
Rearranging leads to the final solution:
\[ 3y = y + x^2 + x \]
Thus, the final answer is:
\[ 3x^2 y = y + 2x \]