A third order linear, homogeneous DE whose general solution is  is: y(t)= c1e^(t) + c2e^(2t) = c3e^(3t) [Hint: The general solution implies that r=1,2 and 3 are the roots of the characteristic equation of the DE. Hence r-1, r-2 and r-3 are the factors of the characteristic equation.]    A. none of these  B. y'''-6y''-11y'-6y=0  C. y'''+6y''-11y'-6y=0  D. y'''+6y''+11y'-6y=0  E.  y'''-6y''+11y'-6y=

Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 12CR
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(a)A third order linear, homogeneous DE whose general solution is  is: y(t)= c1e^(t) + c2e^(2t) = c3e^(3t)

[Hint: The general solution implies that r=1,2 and 3 are the roots of the characteristic equation of the DE. Hence r-1, r-2 and r-3 are the factors of the characteristic equation.]
 

 A. none of these
 B. y'''-6y''-11y'-6y=0
 C. y'''+6y''-11y'-6y=0
 D. y'''+6y''+11y'-6y=0
 E.  y'''-6y''+11y'-6y=                                                                                                                                                                                                                 
 (b) Solving the DE;  dx/dt = (t+x)/t, t>0
 x(t) = Int^(t) +Ct

 with the homogeneous method yields
 where C  is an arbitrary constant.
True or False?
(c)The general solution to the DE in the initial value problem (IVP)
 
y'''+8y'+16y = 0
y(0)= 1
y'(0)=4

.
y(x)=C1xe^(-4x) +Ce^(-4x) 


Imposing the initial conditions, the values of the constants C1  and  C2 are
 

C1=  

C2=

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