Suppose y'+y=0. Which of the following is a possibility for y=f(x)

Options

A. y=tanx
B. y=secx
C. y=sinx
D. `y=1/x, xne0`

Answer

To solve the differential equation \( y' + y = 0 \) and determine which of the given options for \( y = f(x) \) is a possibility, we will analyze each option step by step.

### Step 1: Analyze Option 1: \( y = \tan x \)

1. **Find the derivative**:
\[ y' = \frac{d}{dx}(\tan x) = \sec^2 x \]

2. **Substitute into the equation**:
\[ y' + y = \sec^2 x + \tan x \]

3. **Set the equation to zero**:
\[ \sec^2 x + \tan x = 0 \]

4. **Rewrite in terms of sine and cosine**:
\[ \frac{1}{\cos^2 x} + \frac{\sin x}{\cos x} = 0 \]
\[ \frac{1 + \sin x \cos x}{\cos^2 x} = 0 \]

5. **Solve for \( \sin x \cos x \)**:
\[ 1 + \sin x \cos x = 0 \implies \sin x \cos x = -1 \]
This is impossible since \( \sin x \cos x \) can only take values in the range \([-1/2, 1/2]\).

**Conclusion**: Option 1 is **not possible**.

### Step 2: Analyze Option 2: \( y = \sec x \)

1. **Find the derivative**:
\[ y' = \frac{d}{dx}(\sec x) = \sec x \tan x \]

2. **Substitute into the equation**:
\[ y' + y = \sec x \tan x + \sec x \]

3. **Factor out \( \sec x \)**:
\[ \sec x (\tan x + 1) = 0 \]

4. **Set the equation to zero**:
\[ \tan x + 1 = 0 \implies \tan x = -1 \]
This is possible for values of \( x \) such as \( x = \frac{3\pi}{4} + n\pi \).

**Conclusion**: Option 2 is **possible**.

### Step 3: Analyze Option 3: \( y = \sin x \)

1. **Find the derivative**:
\[ y' = \frac{d}{dx}(\sin x) = \cos x \]

2. **Substitute into the equation**:
\[ y' + y = \cos x + \sin x \]

3. **Set the equation to zero**:
\[ \cos x + \sin x = 0 \implies \sin x = -\cos x \]

4. **Solve for \( x \)**:
\[ \tan x = -1 \]
This is possible for values of \( x \) such as \( x = \frac{3\pi}{4} + n\pi \).

**Conclusion**: Option 3 is **possible**.

### Step 4: Analyze Option 4: \( y = \frac{1}{x} \)

1. **Find the derivative**:
\[ y' = \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2} \]

2. **Substitute into the equation**:
\[ y' + y = -\frac{1}{x^2} + \frac{1}{x} \]

3. **Set the equation to zero**:
\[ -\frac{1}{x^2} + \frac{1}{x} = 0 \]

4. **Solve for \( x \)**:
\[ \frac{1}{x} = \frac{1}{x^2} \implies x = 1 \]
This is possible at \( x = 1 \).

**Conclusion**: Option 4 is **possible**.

### Final Conclusion:
The possible options for \( y = f(x) \) are **Option 2, Option 3, and Option 4**.

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