If `(dy)/dx = (xy)/(x^2 + y^2), y(1) = 1` and `y(x) = e` then `x =`

Options

A. `(sqrt3/2)e`
B. `sqrt3e`
C. `sqrt2e`
D. `e/sqrt2`

Answer

To solve the differential equation \(\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\) with the initial condition \(y(1) = 1\) and to find \(x\) when \(y(x) = e\), we can follow these steps:

### Step 1: Identify the Type of Differential Equation
The given differential equation is homogeneous because both the numerator and denominator are of the same degree (degree 2).

### Step 2: Substitute \(y = ax\)
We can use the substitution \(y = ax\), where \(a\) is a function of \(x\). Then, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = a + x\frac{da}{dx} \]

### Step 3: Substitute into the Differential Equation
Substituting \(y = ax\) into the differential equation:
\[ a + x\frac{da}{dx} = \frac{x(ax)}{x^2 + (ax)^2} \]
This simplifies to:
\[ a + x\frac{da}{dx} = \frac{ax^2}{x^2(1 + a^2)} = \frac{a}{1 + a^2} \]

### Step 4: Rearranging the Equation
Rearranging gives:
\[ x\frac{da}{dx} = \frac{a}{1 + a^2} - a \]
This can be simplified to:
\[ x\frac{da}{dx} = a\left(\frac{1 - a^2}{1 + a^2}\right) \]

### Step 5: Separate Variables
Separating the variables:
\[ \frac{1 + a^2}{a(1 - a^2)} da = -\frac{1}{x} dx \]

### Step 6: Integrate Both Sides
Integrating both sides:
\[ \int \left(\frac{1}{a} + \frac{a}{1 - a^2}\right) da = -\int \frac{1}{x} dx \]
The left side integrates to:
\[ \ln |a| - \frac{1}{2} \ln |1 - a^2| = -\ln |x| + C \]

### Step 7: Simplify the Equation
This can be rewritten as:
\[ \ln |a| - \frac{1}{2} \ln |1 - a^2| + \ln |x| = C \]

### Step 8: Substitute Back \(a = \frac{y}{x}\)
Substituting back \(a = \frac{y}{x}\):
\[ \ln \left|\frac{y}{x}\right| - \frac{1}{2} \ln |1 - \left(\frac{y}{x}\right)^2| + \ln |x| = C \]
This simplifies to:
\[ \ln |y| = C + \frac{1}{2} \ln |1 - \frac{y^2}{x^2}| \]

### Step 9: Use Initial Condition to Find \(C\)
Using the initial condition \(y(1) = 1\):
\[ \ln |1| = C + \frac{1}{2} \ln |1 - 1| \Rightarrow C = 0 \]

### Step 10: Substitute \(C\) Back
The equation becomes:
\[ \ln |y| = \frac{1}{2} \ln |1 - \frac{y^2}{x^2}| \]

### Step 11: Solve for \(y = e\)
Now we need to find \(x\) when \(y = e\):
\[ \ln |e| = \frac{1}{2} \ln |1 - \frac{e^2}{x^2}| \]
This simplifies to:
\[ 1 = \frac{1}{2} \ln |1 - \frac{e^2}{x^2}| \]
Multiplying both sides by 2:
\[ 2 = \ln |1 - \frac{e^2}{x^2}| \]
Exponentiating gives:
\[ e^2 = 1 - \frac{e^2}{x^2} \]
Rearranging leads to:
\[ \frac{e^2}{x^2} = 1 - e^2 \]
Thus:
\[ x^2 = \frac{e^2}{1 - e^2} \]

### Step 12: Final Result
Taking the square root:
\[ x = \pm \frac{e}{\sqrt{1 - e^2}} \]
Since we only consider positive \(x\):
\[ x = \frac{e}{\sqrt{1 - e^2}} \]