Solve `(x^(3)+x^(2)+x+1)(dy)/(dx) =2x^(2)+x, " given that " y=1 " when " x =0.`

Explanation

To solve the differential equation \((x^{3}+x^{2}+x+1)\frac{dy}{dx} = 2x^{2}+x\) given that \(y=1\) when \(x=0\), we will follow these steps:

### Step 1: Rewrite the Equation
First, we can rewrite the equation in a more manageable form:
\[ \frac{dy}{dx} = \frac{2x^{2}+x}{x^{3}+x^{2}+x+1} \]

### Step 2: Factor the Denominator
Next, we will factor the denominator \(x^{3}+x^{2}+x+1\):
\[ x^{3}+x^{2}+x+1 = (x^{2}+1)(x+1) \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{2x^{2}+x}{(x^{2}+1)(x+1)} \]

### Step 3: Partial Fraction Decomposition
Now, we will perform partial fraction decomposition on the right-hand side:
\[ \frac{2x^{2}+x}{(x^{2}+1)(x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^{2}+1} \]
Multiplying through by the denominator \((x^{2}+1)(x+1)\) gives:
\[ 2x^{2}+x = A(x^{2}+1) + (Bx+C)(x+1) \]
Expanding the right-hand side:
\[ 2x^{2}+x = Ax^{2} + A + Bx^{2} + Bx + Cx + C \]
Combining like terms:
\[ 2x^{2}+x = (A+B)x^{2} + (B+C)x + (A+C) \]

### Step 4: Set Up the System of Equations
Now we can set up a system of equations by equating coefficients:
1. \(A + B = 2\)
2. \(B + C = 1\)
3. \(A + C = 0\)

### Step 5: Solve the System of Equations
From equation (3), we have \(C = -A\). Substituting \(C\) into equation (2):
\[ B - A = 1 \implies B = A + 1 \]
Now substituting \(B\) into equation (1):
\[ A + (A + 1) = 2 \implies 2A + 1 = 2 \implies 2A = 1 \implies A = \frac{1}{2} \]
Then, substituting \(A\) back to find \(B\) and \(C\):
\[ B = \frac{1}{2} + 1 = \frac{3}{2}, \quad C = -\frac{1}{2} \]

### Step 6: Substitute Back into the Partial Fractions
Now substituting back, we have:
\[ \frac{dy}{dx} = \frac{1/2}{x+1} + \frac{3/2x - 1/2}{x^{2}+1} \]

### Step 7: Integrate Both Sides
Now, we integrate both sides:
\[ dy = \left(\frac{1/2}{x+1} + \frac{3/2x - 1/2}{x^{2}+1}\right)dx \]
Integrating gives:
\[ y = \frac{1}{2} \ln|x+1| + \frac{3}{2} \ln|x^{2}+1| - \frac{1}{2} \tan^{-1}(x) + C \]

### Step 8: Apply the Initial Condition
Using the initial condition \(y(0) = 1\):
\[ 1 = \frac{1}{2} \ln(1) + \frac{3}{2} \ln(1) - \frac{1}{2} \tan^{-1}(0) + C \]
Since \(\ln(1) = 0\) and \(\tan^{-1}(0) = 0\), we have:
\[ 1 = C \implies C = 1 \]

### Final Solution
Thus, the final solution is:
\[ y = \frac{1}{2} \ln|x+1| + \frac{3}{2} \ln|x^{2}+1| - \frac{1}{2} \tan^{-1}(x) + 1 \]