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: \[ y''' - 4y'' + 4y' = 0. \]