Question

The differential equation formed by eliminating \( a \) and \( b \) from the equation \[ y = ae^{2x} + bxe^{2x} \] is:

• A. \( y''' - 4y'' - 4y' = 0 \)
• B. \( y''' + 4y'' = 0 \)
• C. \( y''' - 4y' = 0 \)
• D. \( y''' - 4y'' + 4y' = 0 \)

Hint:

To eliminate arbitrary constants in a function, differentiate successively until the number of equations matches the number of constants.

Answer 1

The Correct Option is D

Step 1: Given function and differentiation
The given function is: \[ y = ae^{2x} + bxe^{2x}. \] Step 2: First derivative
Differentiating with respect to \( x \): \[ y' = a(2e^{2x}) + b \left( e^{2x} + 2x e^{2x} \right). \] \[ = 2ae^{2x} + b e^{2x} + 2b x e^{2x}. \] Step 3: Second derivative
Differentiating again: \[ y'' = 4ae^{2x} + 2b e^{2x} + 2b e^{2x} + 4b x e^{2x}. \] \[ = 4ae^{2x} + 4b e^{2x} + 4b x e^{2x}. \] Step 4: Third derivative
Differentiating once more: \[ y''' = 8ae^{2x} + 8b e^{2x} + 4b e^{2x} + 8b x e^{2x}. \] \[ = 8ae^{2x} + 12b e^{2x} + 8b x e^{2x}. \] Step 5: Eliminating \( a \) and \( b \)
Using the equations: \[ y = ae^{2x} + bxe^{2x}, \] \[ y' = 2ae^{2x} + b e^{2x} + 2b x e^{2x}, \] \[ y'' = 4ae^{2x} + 4b e^{2x} + 4b x e^{2x}, \] \[ y''' = 8ae^{2x} + 12b e^{2x} + 8b x e^{2x}. \] Solving, we obtain: \[ y''' - 4y'' + 4y' = 0. \] Step 6: Conclusion
Thus, the correct answer is: \[ y''' - 4y'' + 4y' = 0. \]

Answer 2

To solve the problem, we are going to eliminate the constants \( a \) and \( b \) from the equation:

\( y = ae^{2x} + bxe^{2x} \)

We will find higher derivatives of \( y \) and use them to form a differential equation:

Step 1: Differentiate \( y \) with respect to \( x \).

The first derivative is:

\( y' = \frac{d}{dx}(ae^{2x} + bxe^{2x}) = 2ae^{2x} + b(e^{2x} + 2xe^{2x}) = (2a + b)e^{2x} + 2bxe^{2x} \)

Step 2: Differentiate \( y' \) with respect to \( x \).

The second derivative is:

\( y'' = \frac{d}{dx}((2a + b)e^{2x} + 2bxe^{2x}) = 2(2a + b)e^{2x} + 2b(e^{2x} + 2xe^{2x}) = (4a + 4b)e^{2x} + 4bxe^{2x} \)

Step 3: Differentiate \( y'' \) with respect to \( x \).

The third derivative is:

\( y''' = \frac{d}{dx}((4a + 4b)e^{2x} + 4bxe^{2x}) = 2(4a + 4b)e^{2x} + 4b(e^{2x} + 2xe^{2x}) = (8a + 12b)e^{2x} + 8bxe^{2x} \)

Step 4: Eliminate \( a \) and \( b \).

We have the system of equations:

\( y = ae^{2x} + bxe^{2x} \)

\( y' = (2a + b)e^{2x} + 2bxe^{2x} \)

\( y'' = (4a + 4b)e^{2x} + 4bxe^{2x} \)

\( y''' = (8a + 12b)e^{2x} + 8bxe^{2x} \)

Notice the terms in \( y'''\):

If we substitute and rearrange terms by eliminating \( a \) and \( b \), we arrive at the differential equation for the \( e^{2x} \) terms:

\( y''' - 4y'' + 4y' = 0 \)

Thus, the correct differential equation formed is:

\( y''' - 4y'' + 4y' = 0 \)