y = f(x) which has differential equation `y(2xy + e^(x)) dx - e^(x) dy = 0` passing through the point (0,1). Then which of the following is/are true about the function?

Options

A. `x = 1 + sqrt(2)` is point of local maxima
B. `x = 1 - sqrt(2)` is point of local minima
C. `underset(x rarr oo)("lim") f (x) = - oo`
D. `underset(x rarr -oo)("lim") f(x) = 0`

Explanation

To solve the differential equation given by:

\[ y(2xy + e^x)dx - e^x dy = 0 \]

and find the function \( y = f(x) \) that passes through the point \( (0, 1) \), we will proceed step by step.

### Step 1: Rearranging the Equation
We start by rearranging the equation into a more manageable form:

\[ y(2xy + e^x)dx = e^x dy \]

Dividing both sides by \( e^x \):

\[ \frac{y(2xy + e^x)}{e^x}dx = dy \]

This simplifies to:

\[ \left( \frac{2xy^2}{e^x} + y \right) dx = dy \]

### Step 2: Separating Variables
We can rewrite this as:

\[ dy = \left( \frac{2xy^2}{e^x} + y \right) dx \]

This implies:

\[ \frac{dy}{dx} = \frac{2xy^2}{e^x} + y \]

### Step 3: Introducing a New Variable
Let \( V = \frac{1}{y} \). Then, we have:

\[ y = \frac{1}{V} \quad \text{and} \quad \frac{dy}{dx} = -\frac{1}{V^2} \frac{dV}{dx} \]

Substituting these into the equation gives:

\[ -\frac{1}{V^2} \frac{dV}{dx} = \frac{2x}{e^x V^2} + \frac{1}{V} \]

Multiplying through by \( -V^2 \):

\[ \frac{dV}{dx} = -\left( \frac{2x}{e^x} + V \right) \]

### Step 4: Solving the Linear Differential Equation
This is a linear first-order differential equation in \( V \):

\[ \frac{dV}{dx} + V = -\frac{2x}{e^x} \]

The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int 1 \, dx} = e^x \]

Multiplying through by the integrating factor:

\[ e^x \frac{dV}{dx} + e^x V = -2x \]

### Step 5: Integrating Both Sides
The left-hand side can be rewritten as:

\[ \frac{d}{dx}(e^x V) = -2x \]

Integrating both sides:

\[ e^x V = -2 \int x \, dx = -2 \left( \frac{x^2}{2} \right) + C = -x^2 + C \]

Thus,

\[ e^x V = -x^2 + C \]

### Step 6: Back Substituting for \( y \)
Substituting back for \( V \):

\[ V = \frac{1}{y} = \frac{-x^2 + C}{e^x} \]

So,

\[ y = \frac{e^x}{-x^2 + C} \]

### Step 7: Finding the Constant \( C \)
Using the point \( (0, 1) \):

\[ 1 = \frac{e^0}{-0^2 + C} \implies 1 = \frac{1}{C} \implies C = 1 \]

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

\[ y = \frac{e^x}{1 - x^2} \]

### Step 8: Analyzing the Function
Now we analyze the function \( y = \frac{e^x}{1 - x^2} \):

1. **Local Maxima and Minima**: To find local maxima and minima, we differentiate \( y \) and set the derivative to zero.
2. **Behavior as \( x \to \infty \)**: As \( x \to \infty \), \( y \to -\infty \).
3. **Behavior as \( x \to -\infty \)**: As \( x \to -\infty \), \( y \to 0 \).