The radius of `y(sqrt(8)) + (17)/(9)` if `(1+x^(2))(dy)/(dx) = x(1-y), y(0) = (4)/(3)` is ________

Options

A. 2
B. 3
C. 5
D. None of these

Answer

To solve the differential equation given by

\[ (1 + x^2) \frac{dy}{dx} = x(1 - y) \]

with the initial condition \( y(0) = \frac{4}{3} \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation into a more manageable form:

\[ \frac{dy}{1 - y} = -\frac{x}{1 + x^2} dx \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side becomes:

\[ \int \frac{dy}{1 - y} = -\ln|1 - y| \]

For the right side, we can use the substitution \( a = 1 + x^2 \), which gives \( da = 2x dx \) or \( dx = \frac{da}{2x} \). Thus, we have:

\[ \int -\frac{x}{1 + x^2} dx = -\frac{1}{2} \ln(1 + x^2) \]

So, the integration gives us:

\[ -\ln|1 - y| = -\frac{1}{2} \ln(1 + x^2) + C \]

### Step 3: Simplifying the Equation
By exponentiating both sides, we can express this as:

\[ |1 - y| = \frac{C}{\sqrt{1 + x^2}} \]

### Step 4: Applying the Initial Condition
Using the initial condition \( y(0) = \frac{4}{3} \):

\[ |1 - \frac{4}{3}| = \frac{C}{\sqrt{1 + 0^2}} \implies |-\frac{1}{3}| = C \implies C = \frac{1}{3} \]

### Step 5: Final Form of the Equation
Substituting \( C \) back into the equation, we have:

\[ 1 - y = \frac{1/3}{\sqrt{1 + x^2}} \implies y = 1 - \frac{1/3}{\sqrt{1 + x^2}} \]

### Step 6: Finding \( y(\sqrt{8}) \)
Now, we need to find \( y(\sqrt{8}) \):

\[ y(\sqrt{8}) = 1 - \frac{1/3}{\sqrt{1 + (\sqrt{8})^2}} = 1 - \frac{1/3}{\sqrt{1 + 8}} = 1 - \frac{1/3}{\sqrt{9}} = 1 - \frac{1/3}{3} = 1 - \frac{1}{9} = \frac{8}{9} \]

### Step 7: Adding \( \frac{17}{9} \)
Finally, we compute:

\[ y(\sqrt{8}) + \frac{17}{9} = \frac{8}{9} + \frac{17}{9} = \frac{25}{9} \]

### Step 8: Finding the Radius
The radius is the value of \( y(\sqrt{8}) + \frac{17}{9} \):

\[ \text{Radius} = \frac{25}{9} \]

### Final Answer
The radius of \( y(\sqrt{8}) + \frac{17}{9} \) is \( \frac{25}{9} \).

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