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Answer:
The correct answer is:
C. [tex]\( y^{(4)} + 65y'' + 64y = 0 \)[/tex]
Step-by-step explanation:
To find a homogeneous linear differential equation with constant coefficients whose general solution is [tex]\( y = C_1 \cos(x) + C_2 \sin(x) + C_3 \cos(8x) + C_4 \sin(8x) \),[/tex] we must determine the characteristic equation corresponding to this solution.
Given the solution components:
[tex]1. \( \cos(x) \) and \( \sin(x) \)\\2. \( \cos(8x) \) and \( \sin(8x) \)[/tex]
The characteristic roots corresponding to these solutions are:
[tex]1. \( \cos(x) \) and \( \sin(x) \) imply roots \( \pm i \)\\2. \( \cos(8x) \) and \( \sin(8x) \) imply roots \( \pm 8i \)[/tex]
Thus, the characteristic polynomial must have roots [tex]\( \pm I \) and \( \pm 8i \). This gives us the characteristic equation:\[ (r - i)(r + i)(r - 8i)(r + 8i) = 0 \][/tex]
The characteristic polynomial corresponding to these roots is:
[tex]\[ (r^2 + 1)(r^2 + 64) = 0 \][/tex]
Expanding this product, we get:
[tex]\[ (r^2 + 1)(r^2 + 64) = r^4 + 64r^2 + r^2 + 64 = r^4 + 65r^2 + 64 \][/tex]
Therefore, the corresponding differential equation is:
[tex]\[ y^{(4)} + 65y'' + 64y = 0 \][/tex]
So, the correct answer is:
C. [tex]\( y^{(4)} + 65y'' + 64y = 0 \)[/tex]
Homogeneous linear differential equation with constant coefficients D. y⁽⁴⁾ + 65y'' + 64y = 0.
To find a homogeneous linear differential equation with constant coefficients whose general solution is given by:
y = C₁ cos(x) + C₂ sin(x) + C₃ cos(8x) + C₄ sin(8x),
The corresponding characteristic equation will then be:
Expanding this, we get:
This can be rewritten as:
The corresponding differential equation with these characteristic roots is: