Solution of the differential equation `(x^(2)-y^(2))dx+2xy dy=0` is

Options

A. `x^(2)-y^(2)=cx`
B. `x^(2)+y^(2)=cx`
C. `x^(2)-y^(2)=cx^(2)`
D. `x^(2)+y^(2)=cx^(2)`

Answer

To solve the differential equation \((x^{2} - y^{2})dx + 2xy dy = 0\), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation into a more manageable form. We can express it as:
\[ 2xy \, dy = - (x^{2} - y^{2}) \, dx \]
Dividing both sides by \(2xy\) gives:
\[ dy = -\frac{x^{2} - y^{2}}{2xy} \, dx \]

### Step 2: Substituting \(y = vx\)
Next, we use the substitution \(y = vx\), where \(v\) is a function of \(x\). Thus, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting \(y\) and \(\frac{dy}{dx}\) into our equation gives:
\[ v + x \frac{dv}{dx} = -\frac{x^{2} - (vx)^{2}}{2x(vx)} = -\frac{x^{2} - v^{2}x^{2}}{2vx^{2}} = -\frac{(1 - v^{2})}{2v} \]

### Step 3: Rearranging the Equation
Now we have:
\[ v + x \frac{dv}{dx} = -\frac{(1 - v^{2})}{2v} \]
Rearranging this gives:
\[ x \frac{dv}{dx} = -\frac{(1 - v^{2})}{2v} - v \]
\[ x \frac{dv}{dx} = -\frac{(1 - v^{2}) + 2v^{2}}{2v} = -\frac{(1 + v^{2})}{2v} \]

### Step 4: Separating Variables
We can separate variables:
\[ 2v \, dv = -\frac{(1 + v^{2})}{x} \, dx \]
This leads to:
\[ \frac{2v}{1 + v^{2}} \, dv = -\frac{1}{x} \, dx \]

### Step 5: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{2v}{1 + v^{2}} \, dv = -\int \frac{1}{x} \, dx \]
The left side integrates to:
\[ \ln(1 + v^{2}) = -\ln(x) + C \]
where \(C\) is the constant of integration.

### Step 6: Exponentiating Both Sides
Exponentiating both sides gives:
\[ 1 + v^{2} = \frac{C}{x} \]

### Step 7: Substituting Back for \(v\)
Recalling that \(v = \frac{y}{x}\), we substitute back:
\[ 1 + \left(\frac{y}{x}\right)^{2} = \frac{C}{x} \]
Multiplying through by \(x^{2}\) gives:
\[ x^{2} + y^{2} = Cx \]

### Final Solution
Thus, the solution to the differential equation is:
\[ x^{2} + y^{2} = Cx \]