Please show the work - Thanks! Consider the IVP for the function y given by y\"

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Please show the work - Thanks!

Consider the IVP for the function y given by y" + 12y' + 32y = 0, y(0) = -3, y'(0) = -2. (2/10) Find the roots of the characteristic polynomial, r+ and r-, larger and smaller or equal or conjugate, respectively, r+ = r- = (2/10) Find real-valued fundamental solutions to the differential equation, y+(t) = y-(t) = (2/10) Use the fundamental solutions in (b) to find the general solution to the differential equation, (use constant names c and d to construct the general solution), ygen (t) = (4/10) Find the solution y of the initial value problem, y(t) =

Explanation / Answer

The Characteristic polynomial is D^2 + 12 + 32 = 0

This factors as (D + 4)(D+8) = 0, which means roots are -8 and -4

a)

r+ = -4

r- = -8

Thus, associate with r+ = -4 is the fundamental solution

b)

y+(t) = e^-4t and y-(t) = e^-8t

Thus,

c)

ygen(t) = ce^-4t + de^-8t

Then,

d) as y(0) = -3 and y'(0) = -2

From ygent(t)) = ce^-4t + de^-8t

ygen(0) = ce^-4*0 + de^-8*0 = ce^0+de^0 = c*1 + d*1 = c + d

Thus, as y(0) = -3, c + d = -3

y'(t) = -4ce^-4t -8de^-8t

Thus, y'(0) = -4ce^-4*0 -8de^-8*0 = -4ce^0 -8de^0 = -4c*1 -8d*1 = -4c - 8d

As y'0) = -2,

-4c - 8d = -2

Thus, we have the 2 equations

c + d = -3

-4c - 8d = -2

Multiplying the top equation by 4

4c + 4d = -12

Adding the 2 equations,

-4d = -14

d = 7/2

Then, c + d = -3, so

c + 7/2 = -3

c = -3 - 7/2

c = -13/2

Thus, the general solution is

ce^-4t + de^-8t =

-13/2e^-4t + 7/2e^-8t

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