Question

If \( y = A + Be^x \), then prove that \( \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0 \), where A and B are constants.

Hint:

An alternative approach is to notice that from the first and second derivatives, we have \( \frac{d^2y}{dx^2} = Be^x \) and \( \frac{dy}{dx} = Be^x \). This directly implies that \( \frac{d^2y}{dx^2} = \frac{dy}{dx} \), which rearranges to the required equation.

Answer 1

Step 1: Understanding the Concept:
To prove that the given function satisfies the differential equation, we need to find the first and second derivatives of the function and substitute them into the equation to show that it holds true.
Step 2: Key Formula or Approach:
1. Find the first derivative, \( \frac{dy}{dx} \).
2. Find the second derivative, \( \frac{d^2y}{dx^2} \), by differentiating the first derivative.
3. Substitute the expressions for the derivatives into the left-hand side of the given differential equation and show that it simplifies to 0.
Step 3: Detailed Explanation or Calculation:
The given function is \( y = A + Be^x \), where A and B are constants.
First Derivative:
\[ \frac{dy}{dx} = \frac{d}{dx}(A + Be^x) = \frac{d}{dx}(A) + B\frac{d}{dx}(e^x) = 0 + Be^x = Be^x \] Second Derivative:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(Be^x) = B\frac{d}{dx}(e^x) = Be^x \] Now, substitute these derivatives into the expression \( \frac{d^2y}{dx^2} - \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} - \frac{dy}{dx} = (Be^x) - (Be^x) = 0 \] Step 4: Final Answer:
We have shown that \( \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0 \). Hence proved.