`x=x(t)` solution of `(t+1)dx=[2x+(t+1)^3]dtx(0)=2` then x(1)=

Options

A. 6
B. 8
C. 12
D. 10

Answer

To solve the given differential equation \((t+1)dx = [2x + (t+1)^3]dt\) with the initial condition \(x(0) = 2\), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start by rewriting the equation in a more standard form:
\[ dx = \frac{2x + (t+1)^3}{t+1} dt \]
This simplifies to:
\[ dx = \left(\frac{2x}{t+1} + (t+1)^2\right) dt \]

### Step 2: Rearranging into Linear Form
We can rearrange this into the standard form of a linear differential equation:
\[ \frac{dx}{dt} - \frac{2}{t+1} x = (t+1)^2 \]

### Step 3: Identify the Integrating Factor
The integrating factor \(I(t)\) is given by:
\[ I(t) = e^{\int -\frac{2}{t+1} dt} = e^{-2 \ln(t+1)} = (t+1)^{-2} \]

### Step 4: Multiply the Equation by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor:
\[ (t+1)^{-2} \frac{dx}{dt} - \frac{2}{(t+1)^3} x = (t+1)^0 \]

### Step 5: Rewrite the Left Side
The left side can be rewritten as the derivative of a product:
\[ \frac{d}{dt}\left(x(t+1)^{-2}\right) = 1 \]

### Step 6: Integrate Both Sides
Integrating both sides with respect to \(t\):
\[ \int \frac{d}{dt}\left(x(t+1)^{-2}\right) dt = \int 1 dt \]
This gives:
\[ x(t+1)^{-2} = t + C \]
where \(C\) is the constant of integration.

### Step 7: Solve for \(x\)
Now, we can solve for \(x\):
\[ x = (t + C)(t + 1)^2 \]

### Step 8: Apply the Initial Condition
Using the initial condition \(x(0) = 2\):
\[ 2 = (0 + C)(0 + 1)^2 \implies 2 = C \]
Thus, \(C = 2\).

### Step 9: Write the Particular Solution
Substituting \(C\) back into the equation for \(x\):
\[ x = (t + 2)(t + 1)^2 \]

### Step 10: Evaluate \(x(1)\)
Now we need to find \(x(1)\):
\[ x(1) = (1 + 2)(1 + 1)^2 = 3 \cdot 2^2 = 3 \cdot 4 = 12 \]

### Final Answer
Thus, the value of \(x(1)\) is:
\[ \boxed{12} \]