Let y=y(x) be the solution of the differential equation `(x+1)y'-y= e^(3x) (x+1)^(2) ` with y(0) `= 1/3` Then, the point x `= -4/3` for the curve `y=y(x)` is :

Options

A. not a critical point
B. a point of local minima
C. a point of local maxima
D. a point of inflection

Answer

To solve the differential equation given by

\[ (x+1)y' - y = e^{3x}(x+1)^2 \]

with the initial condition \( y(0) = \frac{1}{3} \), we will follow these steps:

### Step 1: Rewrite the Equation
We can rewrite the equation in a more standard form:

\[ y' - \frac{y}{x+1} = e^{3x}(x+1) \]

### Step 2: Identify the Integrating Factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int -\frac{1}{x+1} \, dx} = e^{-\ln|x+1|} = \frac{1}{x+1} \]

### Step 3: Multiply the Equation by the Integrating Factor
Multiplying the entire differential equation by the integrating factor:

\[ \frac{1}{x+1} \left( (x+1)y' - y \right) = \frac{1}{x+1} e^{3x}(x+1)^2 \]

This simplifies to:

\[ y' - \frac{y}{x+1} = e^{3x}(x+1) \]

### Step 4: Solve the Left Side
The left side can be expressed as the derivative of a product:

\[ \frac{d}{dx} \left( \frac{y}{x+1} \right) = e^{3x}(x+1) \]

### Step 5: Integrate Both Sides
Integrating both sides with respect to \( x \):

\[ \int \frac{d}{dx} \left( \frac{y}{x+1} \right) \, dx = \int e^{3x}(x+1) \, dx \]

The left side simplifies to:

\[ \frac{y}{x+1} = \int e^{3x}(x+1) \, dx \]

### Step 6: Solve the Right Side Integral
To solve the integral on the right side, we can use integration by parts. Let:

- \( u = x + 1 \)
- \( dv = e^{3x}dx \)

Then, \( du = dx \) and \( v = \frac{1}{3} e^{3x} \).

Using integration by parts:

\[ \int e^{3x}(x+1) \, dx = \frac{1}{3} e^{3x}(x+1) - \int \frac{1}{3} e^{3x} \, dx \]

Calculating the integral:

\[ = \frac{1}{3} e^{3x}(x+1) - \frac{1}{9} e^{3x} + C \]

### Step 7: Substitute Back
Substituting back, we have:

\[ \frac{y}{x+1} = \frac{1}{3} e^{3x}(x+1) - \frac{1}{9} e^{3x} + C \]

Multiplying through by \( x + 1 \):

\[ y = \frac{1}{3} e^{3x}(x+1)^2 - \frac{1}{9} e^{3x}(x+1) + C(x + 1) \]

### Step 8: Use Initial Condition
Using the initial condition \( y(0) = \frac{1}{3} \):

\[ \frac{1}{3} = \frac{1}{3} e^{0}(1)^2 - \frac{1}{9} e^{0}(1) + C(1) \]

This simplifies to:

\[ \frac{1}{3} = \frac{1}{3} - \frac{1}{9} + C \]

Solving for \( C \):

\[ C = \frac{1}{3} - \frac{1}{3} + \frac{1}{9} = \frac{1}{9} \]

### Step 9: Final Solution
Thus, the solution to the differential equation is:

\[ y = \frac{1}{3} e^{3x}(x+1)^2 - \frac{1}{9} e^{3x}(x+1) + \frac{1}{9}(x + 1) \]

### Step 10: Evaluate at \( x = -\frac{4}{3} \)
Now we need to evaluate \( y \) at \( x = -\frac{4}{3} \):

Substituting \( x = -\frac{4}{3} \):

\[ y\left(-\frac{4}{3}\right) = \frac{1}{3} e^{-4}( -\frac{4}{3} + 1)^2 - \frac{1}{9} e^{-4}(-\frac{4}{3} + 1) + \frac{1}{9}(-\frac{4}{3} + 1) \]

Calculating this will give us the value of \( y \) at \( x = -\frac{4}{3} \).