If `(x+x^3)dy=(y+yx^2+x^3)dx and y(1)=0` then `y(2)` is

Options

A. `ln(17/2)`
B. 0
C. `ln(5/2)`
D. `ln(2/5)`

Answer

To solve the differential equation given by \((x + x^3) dy = (y + yx^2 + x^3) dx\) with the initial condition \(y(1) = 0\), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the equation in the standard form:
\[ \frac{dy}{dx} = \frac{y + yx^2 + x^3}{x + x^3} \]

### Step 2: Simplify the Right-Hand Side
We can simplify the right-hand side:
\[ \frac{dy}{dx} = \frac{y(1 + x^2) + x^3}{x(1 + x^2)} \]
This can be separated into two parts:
\[ \frac{dy}{dx} = \frac{y(1 + x^2)}{x(1 + x^2)} + \frac{x^3}{x(1 + x^2)} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y}{x} + \frac{x^2}{1 + x^2} \]

### Step 3: Rearranging the Equation
Rearranging gives us:
\[ \frac{dy}{dx} - \frac{y}{x} = \frac{x^2}{1 + x^2} \]

### Step 4: Identify \(p\) and \(q\)
From the standard form \( \frac{dy}{dx} + p y = q \), we identify:
- \(p = -\frac{1}{x}\)
- \(q = \frac{x^2}{1 + x^2}\)

### Step 5: Find the Integrating Factor
The integrating factor \(I\) is given by:
\[ I = e^{\int p \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln |x|} = \frac{1}{x} \]

### Step 6: Multiply the Equation by the Integrating Factor
Multiplying the entire equation by the integrating factor:
\[ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{x^2}{x(1 + x^2)} = \frac{x}{1 + x^2} \]

### Step 7: Rewrite the Left-Hand Side
The left-hand side can be rewritten as:
\[ \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{x}{1 + x^2} \]

### Step 8: Integrate Both Sides
Integrating both sides gives:
\[ \frac{y}{x} = \int \frac{x}{1 + x^2} \, dx + C \]

Using the substitution \(u = 1 + x^2\), we find:
\[ \int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln |1 + x^2| + C \]
Thus,
\[ \frac{y}{x} = \frac{1}{2} \ln(1 + x^2) + C \]

### Step 9: Solve for \(y\)
Multiplying through by \(x\):
\[ y = \frac{x}{2} \ln(1 + x^2) + Cx \]

### Step 10: Apply the Initial Condition
Using the initial condition \(y(1) = 0\):
\[ 0 = \frac{1}{2} \ln(2) + C \cdot 1 \]
Thus,
\[ C = -\frac{1}{2} \ln(2) \]

### Step 11: Substitute \(C\) Back into the Equation
Substituting \(C\) back, we have:
\[ y = \frac{x}{2} \ln(1 + x^2) - \frac{1}{2} \ln(2) \cdot x \]

### Step 12: Find \(y(2)\)
Now, substituting \(x = 2\):
\[ y(2) = \frac{2}{2} \ln(1 + 2^2) - \frac{1}{2} \ln(2) \cdot 2 \]
This simplifies to:
\[ y(2) = \ln(5) - \ln(2) = \ln\left(\frac{5}{2}\right) \]

### Final Answer
Thus, the value of \(y(2)\) is:
\[ \boxed{\ln\left(\frac{5}{2}\right)} \]