Solve the following differential equations
`(x^2-y^2)dx+2xy""dy=0, y(1)=1`.

Answer

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

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

### Step 2: Identifying the type of differential equation
Notice that the equation is homogeneous because the degrees of the numerator and denominator are the same. We can express \(y\) in terms of \(x\) by using the substitution \(v = \frac{y}{x}\), which implies \(y = vx\).

### Step 3: Differentiating using the substitution
Differentiating \(y = vx\) with respect to \(x\) gives:
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Substituting \(y = vx\) into the rearranged equation gives:
\[ \frac{dy}{dx} = -\frac{x^2 - (vx)^2}{2x(vx)} = -\frac{x^2(1 - v^2)}{2vx^2} = -\frac{1 - v^2}{2v} \]

### Step 4: Setting up the equation
Now we have:
\[ v + x\frac{dv}{dx} = -\frac{1 - v^2}{2v} \]
Rearranging gives:
\[ x\frac{dv}{dx} = -\frac{1 - v^2}{2v} - v \]
Combining the terms on the right:
\[ x\frac{dv}{dx} = -\frac{1 - v^2 + 2v^2}{2v} = -\frac{1 + v^2}{2v} \]

### Step 5: Separating variables
We can separate variables:
\[ \frac{2v}{1 + v^2} dv = -\frac{dx}{x} \]

### Step 6: Integrating both sides
Integrating both sides:
\[ \int \frac{2v}{1 + v^2} dv = \int -\frac{dx}{x} \]
The left side integrates to \(\log(1 + v^2)\) and the right side integrates to \(-\log|x| + C\):
\[ \log(1 + v^2) = -\log|x| + C \]

### Step 7: Exponentiating both sides
Exponentiating both sides gives:
\[ 1 + v^2 = \frac{C}{|x|} \]
Substituting back \(v = \frac{y}{x}\):
\[ 1 + \left(\frac{y}{x}\right)^2 = \frac{C}{|x|} \]
Multiplying through by \(x^2\) gives:
\[ x^2 + y^2 = Cx \]

### Step 8: Applying the initial condition
Using the initial condition \(y(1) = 1\):
\[ 1^2 + 1^2 = C \cdot 1 \implies 2 = C \]
Thus, we have:
\[ x^2 + y^2 = 2x \]

### Final Step: Rearranging the equation
Rearranging gives us the final solution:
\[ x^2 + y^2 - 2x = 0 \]