The equation of the curve whose slope at any point is equal to `y+ 2x` and which passes through the origin is

Options

A. `y= 2(x-1)`
B. `y= 2(e^(x)-x-1)`
C. `y= 2(e^(x)-1)`
D. `y= 2(e^(x) x-1)`

Answer

To solve the problem, we need to find the equation of the curve whose slope at any point is given by the expression \( y + 2x \) and which passes through the origin (0, 0).

### Step-by-Step Solution:

1. **Identify the Differential Equation**:
The slope of the curve at any point is given by:
\[ \frac{dy}{dx} = y + 2x \]

2. **Rearrange the Equation**:
Rearranging gives:
\[ \frac{dy}{dx} - y = 2x \]
This is a first-order linear differential equation.

3. **Find the Integrating Factor**:
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int -1 \, dx} = e^{-x} \]

4. **Multiply the Equation by the Integrating Factor**:
Multiply the entire differential equation by \( e^{-x} \):
\[ e^{-x} \frac{dy}{dx} - e^{-x} y = 2x e^{-x} \]

5. **Rewrite the Left Side**:
The left side can be rewritten as:
\[ \frac{d}{dx}(y e^{-x}) = 2x e^{-x} \]

6. **Integrate Both Sides**:
Now, integrate both sides:
\[ \int \frac{d}{dx}(y e^{-x}) \, dx = \int 2x e^{-x} \, dx \]
The left side simplifies to:
\[ y e^{-x} \]
For the right side, we can use integration by parts:
Let \( u = 2x \) and \( dv = e^{-x} dx \).
Then \( du = 2 dx \) and \( v = -e^{-x} \).
Applying integration by parts:
\[ \int 2x e^{-x} \, dx = -2x e^{-x} - \int -2 e^{-x} \, dx = -2x e^{-x} + 2 e^{-x} + C \]

7. **Combine Results**:
Thus, we have:
\[ y e^{-x} = -2x e^{-x} + 2 e^{-x} + C \]

8. **Multiply by \( e^{x} \)**:
To solve for \( y \), multiply through by \( e^{x} \):
\[ y = -2x + 2 + Ce^{x} \]

9. **Apply Initial Condition**:
Since the curve passes through the origin (0, 0), substitute \( x = 0 \) and \( y = 0 \):
\[ 0 = -2(0) + 2 + Ce^{0} \implies 0 = 2 + C \implies C = -2 \]

10. **Final Equation**:
Substitute \( C \) back into the equation:
\[ y = -2x + 2 - 2e^{x} \]

### Final Answer:
The equation of the curve is:
\[ y = -2x + 2 - 2e^{x} \]