`(x^(2)+y^(2))dy=xy dx`. If `y(x_(0))=e, y(1)=1`, then value of `x_(0)=`

Options

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

Answer

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

### Step 1: Rewrite the Differential Equation
We start with the equation:
\[ (x^2 + y^2) dy = xy dx \]
We can rewrite this as:
\[ \frac{dy}{dx} = \frac{xy}{x^2 + y^2} \]

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

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

### Step 4: Separate Variables
Separating variables gives:
\[ \frac{1}{v^2} dv = -\frac{1}{x} dx \]

### Step 5: Integrate Both Sides
Integrating both sides:
\[ \int \frac{1}{v^2} dv = \int -\frac{1}{x} dx \]
This results in:
\[ -\frac{1}{v} = -\ln|x| + C \]
or:
\[ \frac{1}{v} = \ln|x| + C \]

### Step 6: Substitute Back for \(y\)
Since \(v = \frac{y}{x}\), we have:
\[ \frac{x}{y} = \ln|x| + C \implies y = \frac{x}{\ln|x| + C} \]

### Step 7: Use Initial Condition \(y(1) = 1\)
Substituting \(x = 1\) and \(y = 1\):
\[ 1 = \frac{1}{\ln(1) + C} \implies 1 = \frac{1}{0 + C} \implies C = 1 \]
Thus, the equation becomes:
\[ y = \frac{x}{\ln|x| + 1} \]

### Step 8: Use Condition \(y(x_0) = e\)
Now we substitute \(y = e\):
\[ e = \frac{x_0}{\ln|x_0| + 1} \]
This leads to:
\[ e(\ln|x_0| + 1) = x_0 \]
Rearranging gives:
\[ e \ln|x_0| + e - x_0 = 0 \]

### Step 9: Solve for \(x_0\)
This is a transcendental equation, and we can use numerical methods or trial and error to find \(x_0\). We can test values:
- For \(x_0 = 3\):
\[ e \ln(3) + e - 3 \approx 2.718 \cdot 1.0986 + 2.718 - 3 \approx 2.98 - 3 \approx -0.02 \]
- For \(x_0 = 4\):
\[ e \ln(4) + e - 4 \approx 2.718 \cdot 1.3863 + 2.718 - 4 \approx 3.77 - 4 \approx -0.23 \]
- For \(x_0 = 2\):
\[ e \ln(2) + e - 2 \approx 2.718 \cdot 0.6931 + 2.718 - 2 \approx 1.88 + 2.718 - 2 \approx 2.6 \]

By further refining, we find:
- For \(x_0 = 2.5\):
\[ e \ln(2.5) + e - 2.5 \approx 2.718 \cdot 0.9163 + 2.718 - 2.5 \approx 2.49 - 2.5 \approx -0.01 \]

Thus, we find that \(x_0\) is approximately \(2.5\).

### Final Answer
The value of \(x_0\) is approximately \(2.5\).