D.E., having the solution `y=c_(1)+c_(2)e^(3x)`, is A)`y_(2)=3y` B)`y_(2)=3y_(1)` C)`y_(3)+3y_(1)=0` D)`y_(2)+3y=0`

Options

A. `y_(2)=3y`
B. `y_(2)=3y_(1)`
C. `y_(3)+3y_(1)=0`
D. `y_(2)+3y=0`

Answer

To find the differential equation corresponding to the given solution \( y = c_1 + c_2 e^{3x} \), we will follow these steps:

### Step 1: Differentiate the given function once
Given:
\[ y = c_1 + c_2 e^{3x} \]
Differentiating with respect to \( x \):
\[ y' = \frac{dy}{dx} = 0 + c_2 \cdot \frac{d}{dx}(e^{3x}) = c_2 \cdot 3e^{3x} = 3c_2 e^{3x} \]

### Step 2: Differentiate the first derivative to find the second derivative
Now, we differentiate \( y' \):
\[ y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}(3c_2 e^{3x}) = 3c_2 \cdot \frac{d}{dx}(e^{3x}) = 3c_2 \cdot 3e^{3x} = 9c_2 e^{3x} \]

### Step 3: Relate \( y'' \) to \( y' \)
From our expressions, we have:
\[ y' = 3c_2 e^{3x} \]
\[ y'' = 9c_2 e^{3x} \]
Now, we can express \( y'' \) in terms of \( y' \):
\[ y'' = 3y' \]

### Step 4: Relate \( y'' \) to \( y \)
We also have:
\[ y = c_1 + c_2 e^{3x} \]
From this, we can express \( c_2 e^{3x} \) as:
\[ c_2 e^{3x} = y - c_1 \]
Substituting this into the equation for \( y'' \):
\[ y'' = 9(c_2 e^{3x}) = 9(y - c_1) \]
However, since \( c_1 \) is a constant, we can express the relationship in terms of \( y \) and \( y' \):
\[ y'' - 3y' = 0 \]

### Conclusion
The differential equation derived from the given solution \( y = c_1 + c_2 e^{3x} \) is:
\[ y'' - 3y' = 0 \]
This corresponds to option B: \( y_2 = 3y_1 \).