If the solution `y = y(x)` of the differential equation `(x^4 + 2x^3 + 3x^2 + 2x + 2)dy – (2x^2 + 2x + 3)dx = 0` satisfy `y(-1) = -pi/4`, then `y(0)` is equal to

Options

A. `-pi/(12)`
B. 0
C. `pi/2`
D. `pi/4`

Answer

To solve the given differential equation and find \( y(0) \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
The given differential equation is:
\[ (x^4 + 2x^3 + 3x^2 + 2x + 2) dy - (2x^2 + 2x + 3) dx = 0 \]
We can rearrange it to separate variables:
\[ \frac{dy}{dx} = \frac{2x^2 + 2x + 3}{x^4 + 2x^3 + 3x^2 + 2x + 2} \]

### Step 2: Integrate Both Sides
Now we need to integrate both sides:
\[ y = \int \frac{2x^2 + 2x + 3}{x^4 + 2x^3 + 3x^2 + 2x + 2} \, dx \]

### Step 3: Simplify the Integral
To simplify the integral, we can perform polynomial long division or partial fraction decomposition. However, in this case, we notice that the denominator can be factored or simplified.

After some algebraic manipulation, we can express the integral as:
\[ y = \int \left( \frac{1}{x^2 + 1} + \frac{1}{(x+1)^2 + 1} \right) dx \]

### Step 4: Solve the Integral
Now we can integrate term by term:
1. The integral of \( \frac{1}{x^2 + 1} \) is \( \tan^{-1}(x) \).
2. The integral of \( \frac{1}{(x+1)^2 + 1} \) is \( \tan^{-1}(x + 1) \).

Thus, we have:
\[ y = \tan^{-1}(x) + \tan^{-1}(x + 1) + C \]

### Step 5: Use the Initial Condition
We are given the initial condition \( y(-1) = -\frac{\pi}{4} \). Substituting \( x = -1 \):
\[ -\frac{\pi}{4} = \tan^{-1}(-1) + \tan^{-1}(0) + C \]
Since \( \tan^{-1}(-1) = -\frac{\pi}{4} \) and \( \tan^{-1}(0) = 0 \), we have:
\[ -\frac{\pi}{4} = -\frac{\pi}{4} + C \implies C = 0 \]

### Step 6: Write the Particular Solution
Thus, the particular solution is:
\[ y = \tan^{-1}(x) + \tan^{-1}(x + 1) \]

### Step 7: Find \( y(0) \)
Now we need to find \( y(0) \):
\[ y(0) = \tan^{-1}(0) + \tan^{-1}(1) = 0 + \frac{\pi}{4} = \frac{\pi}{4} \]

### Final Answer
Thus, the value of \( y(0) \) is:
\[ \boxed{\frac{\pi}{4}} \]