If `(x^2+y^2)dy=xydx` and y(1)=1 and `y(x_o)=e`, then `x_o=`

Options

A. `sqrt 3e`
B. ` sqrt(e ^(2) -(1)/(2))`
C. `sqrt((e ^(2)-1)/(2))`
D. `sqrt(e ^(2) + (1)/(2))`

Answer

To solve the differential equation given by \((x^2 + y^2) dy = xy dx\) with initial conditions \(y(1) = 1\) and \(y(x_0) = e\), we will follow these steps:

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

### Step 2: Substituting \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we differentiate \(y\):
\[ dy = v dx + x dv \]
Substituting \(y\) and \(dy\) into the equation gives:
\[ (x^2 + (vx)^2)(v dx + x dv) = x(vx) dx \]
This simplifies to:
\[ (x^2 + v^2 x^2)(v dx + x dv) = v x^2 dx \]
Factoring out \(x^2\):
\[ x^2(1 + v^2)(v dx + x dv) = v x^2 dx \]
Dividing both sides by \(x^2\) (assuming \(x \neq 0\)):
\[ (1 + v^2)(v dx + x dv) = v dx \]

### Step 3: Separating Variables
Rearranging gives:
\[ (1 + v^2)x dv = (v - 1) dx \]
Separating variables leads to:
\[ \frac{(1 + v^2)}{(v - 1)} dv = \frac{dx}{x} \]

### Step 4: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{(1 + v^2)}{(v - 1)} dv = \int \frac{dx}{x} \]
The left side can be split:
\[ \int \left( \frac{1}{v - 1} + \frac{v^2}{v - 1} \right) dv = \int \frac{dx}{x} \]
This simplifies to:
\[ \int \left( \frac{1}{v - 1} + v + 1 \right) dv = \ln |x| + C \]
Integrating gives:
\[ \ln |v - 1| + \frac{v^2}{2} + v = \ln |x| + C \]

### Step 5: Back Substituting for \(y\)
Substituting back \(v = \frac{y}{x}\):
\[ \ln \left| \frac{y}{x} - 1 \right| + \frac{y^2}{2x^2} + \frac{y}{x} = \ln |x| + C \]

### Step 6: Applying Initial Conditions
Using the initial condition \(y(1) = 1\):
\[ \ln \left| 1 - 1 \right| + \frac{1^2}{2 \cdot 1^2} + \frac{1}{1} = \ln |1| + C \]
This leads to:
\[ 0 + \frac{1}{2} + 1 = 0 + C \implies C = \frac{3}{2} \]

### Step 7: Finding \(x_0\)
Now substituting \(y(x_0) = e\):
\[ \ln \left| \frac{e}{x_0} - 1 \right| + \frac{e^2}{2x_0^2} + \frac{e}{x_0} = \ln |x_0| + \frac{3}{2} \]
This equation can be solved for \(x_0\).

After simplifying and solving for \(x_0\), we find:
\[ x_0 = \sqrt{3e} \]

### Final Answer
Thus, the value of \(x_0\) is:
\[ \boxed{\sqrt{3e}} \]