Solved

What Is the General Solution of the Third-Order Homogeneous Differential y=C1+C2e2tcos(4t)+C3e2tsin(4t) y=C_{1}+C_{2} e^{2 t} \cos (4 t)+C_{3} e^{2 t} \sin (4 t)

Question 43

Multiple Choice

What is the general solution of the third-order homogeneous differential equation  What is the general solution of the third-order homogeneous differential equation   ? A)    y=C_{1}+C_{2} e^{2 t} \cos (4 t) +C_{3} e^{2 t} \sin (4 t)    B)    y=C_{1}+C_{2} e^{2 t} \cos (2 t) +C_{3} e^{2 t} \sin (2 t)    C)    y=C_{1}+C_{2} e^{-2 t} \cos (2 t) +C_{3} e^{-2 t} \sin (2 t)    D)    y=C_{1}+C_{2} e^{4 t} \cos (2 t) +C_{3} e^{4 t_{5} \sin (2 t) }   E)    y=C_{1}+C_{2} e^{-4 t} \cos (2 t) +C_{3} e^{-4 t} \sin (2 t)  ?


A) y=C1+C2e2tcos(4t) +C3e2tsin(4t) y=C_{1}+C_{2} e^{2 t} \cos (4 t) +C_{3} e^{2 t} \sin (4 t)
B) y=C1+C2e2tcos(2t) +C3e2tsin(2t) y=C_{1}+C_{2} e^{2 t} \cos (2 t) +C_{3} e^{2 t} \sin (2 t)
C) y=C1+C2e2tcos(2t) +C3e2tsin(2t) y=C_{1}+C_{2} e^{-2 t} \cos (2 t) +C_{3} e^{-2 t} \sin (2 t)
D) y=C1+C2e4tcos(2t) +C3e4t5sin(2t) y=C_{1}+C_{2} e^{4 t} \cos (2 t) +C_{3} e^{4 t_{5} \sin (2 t) }
E) y=C1+C2e4tcos(2t) +C3e4tsin(2t) y=C_{1}+C_{2} e^{-4 t} \cos (2 t) +C_{3} e^{-4 t} \sin (2 t)

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions