Question

a, b, c, d are arbitrary constants. Then the corresponding differential equation to \( y = ae^x + be^{-x} + c \cos x + d \sin x \) is

• A. \( y^{(4)} = y \)
• B. \( y^{(4)} + y = 0 \)
• C. \( y^{(4)} - y^{(2)} + 1 = 0 \)
• D. \( y^{(4)} + 2y^{(2)} + 1 = 0 \)

Hint:

For a given solution involving exponential and trigonometric functions, the corresponding linear homogeneous differential equation with constant coefficients can be found by determining the roots of the characteristic equation. Exponential terms \( e^{\alpha x} \) correspond to real roots \( r = \alpha \), and terms \( c_1 \cos(\beta x) + c_2 \sin(\beta x) \) correspond to complex conjugate roots \( r = \pm i\beta \). Form the characteristic equation using these roots and then replace \( r^n \) with \( y^{(n)} \) to get the differential equation.

Answer 1

The Correct Option is A

The given function is \( y = ae^x + be^{-x} + c \cos x + d \sin x \).
The characteristic equation corresponding to \( ae^x + be^{-x} \) has roots \( r = 1, -1 \), which gives the factor \( (r - 1)(r + 1) = r^2 - 1 \).
The characteristic equation corresponding to \( c \cos x + d \sin x \) has roots \( r = i, -i \), which gives the factor \( (r - i)(r + i) = r^2 + 1 \).
The overall characteristic equation for the differential equation is the product of these factors: $$ (r^2 - 1)(r^2 + 1) = 0 $$ $$ r^4 - 1 = 0 $$ The corresponding linear homogeneous differential equation with constant coefficients is obtained by replacing \( r^n \) with \( y^{(n)} \): $$ y^{(4)} - y = 0 $$ There seems to be a mismatch with the provided correct answer.
Let's re-check the characteristic roots and the corresponding differential equation.
The part \( ae^x + be^{-x} \) corresponds to roots \( 1, -1 \).
The part \( c \cos x + d \sin x \) corresponds to roots \( i, -i \).
The characteristic equation is \( (r - 1)(r + 1)(r - i)(r + i) = (r^2 - 1)(r^2 + 1) = r^4 - 1 = 0 \).
The corresponding differential equation is \( y^{(4)} - y = 0 \).
If the correct answer is \( y^{(4)} = y \), it implies the characteristic equation is \( r^4 - 1 = 0 \), which matches our derivation.
Final Answer: The final answer is $\boxed{y^{(4)} = y}$