`y=ae^(-(1)/(x))+b` is a solution of `(dy)/(dx)=(y)/(x^(2))` when

Options

A. `a=1, b=0`
B. `a=0, b=2`
C. `a=1, b=1`
D. `a=2, b=2`

Answer

To determine the values of \( a \) and \( b \) such that \( y = ae^{-\frac{1}{x}} + b \) is a solution of the differential equation \( \frac{dy}{dx} = \frac{y}{x^2} \), we will follow these steps:

### Step 1: Differentiate \( y \)
We start by differentiating \( y \) with respect to \( x \).

\[ y = ae^{-\frac{1}{x}} + b \]

Using the chain rule, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = a \cdot \frac{d}{dx}\left(e^{-\frac{1}{x}}\right) + 0 \]

To differentiate \( e^{-\frac{1}{x}} \), we apply the chain rule:

\[ \frac{d}{dx}\left(e^{-\frac{1}{x}}\right) = e^{-\frac{1}{x}} \cdot \frac{d}{dx}\left(-\frac{1}{x}\right) = e^{-\frac{1}{x}} \cdot \left(\frac{1}{x^2}\right) \]

Thus, we have:

\[ \frac{dy}{dx} = a e^{-\frac{1}{x}} \cdot \frac{1}{x^2} \]

### Step 2: Substitute into the Differential Equation
Now we substitute \( y \) and \( \frac{dy}{dx} \) into the differential equation \( \frac{dy}{dx} = \frac{y}{x^2} \).

From our previous calculations:

\[ \frac{dy}{dx} = \frac{a e^{-\frac{1}{x}}}{x^2} \]

And we have:

\[ \frac{y}{x^2} = \frac{ae^{-\frac{1}{x}} + b}{x^2} \]

### Step 3: Set the Two Expressions Equal
We set the two expressions equal to each other:

\[ \frac{a e^{-\frac{1}{x}}}{x^2} = \frac{ae^{-\frac{1}{x}} + b}{x^2} \]

### Step 4: Simplify the Equation
Multiplying both sides by \( x^2 \) (assuming \( x \neq 0 \)) gives:

\[ a e^{-\frac{1}{x}} = ae^{-\frac{1}{x}} + b \]

Subtracting \( ae^{-\frac{1}{x}} \) from both sides results in:

\[ 0 = b \]

### Step 5: Conclusion for \( b \)
From the above, we conclude that:

\[ b = 0 \]

### Step 6: Determine \( a \)
Since \( a \) can be any real number (as it does not affect the equality), we conclude that:

\[ a \in \mathbb{R} \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:

\[ a \text{ can be any real number, and } b = 0 \]