recreating a secound order linear from the solution

[ijd5000]

a)
Find the second order linear equation which has y₁=-e^(6t) - 8e^(-3t) and y₂= 2e^(-3t+5)

b)Find the second order linear which has y= -37 pi te^(-11t) as one of it's solutions.


I'm thinking I have to recreate the characteristic polynomial, but not sure how to deal with the -8e^(-3t) in y₁ and the (-3t+5) from y_{2 .}

 

[Deleted member 4993]

a)
Find the second order linear equation which has y₁=-e^(6t) - 8e^(-3t) and y₂= 2e^(-3t+5)

b)Find the second order linear which has y= -37 pi te^(-11t) as one of it's solutions.


I'm thinking I have to recreate the characteristic polynomial, but not sure how to deal with the -8e^(-3t) in y₁ and the (-3t+5) from y_{2 .}

Differentiate (y1 + y2) twice and look at what relationship is possible.

 

[ijd5000]

what do you mean by relationship?

y"=-36e^(6t)-72e^(-3t)+18e^(-3t+5)

do I call this my r^2?

*edit: forgot to hold shift

 

[Deleted member 4993]

what do you mean by relationship?

y"=-36e^(6t)-72e^9-3t)+18e^(-3t+5) → Those are incorrect

do I call this my r^2?

Set up

M*y" + N*y' + y = 0 as your ODE

solve for M and N

 

[ijd5000]

thank you,

i got

M=-1/18
N=1/6

is this correct?

also what about a coefficient for the y term?

 

[Deleted member 4993]

thank you,

i got

M=-1/18
N=1/6

is this correct?

also what about a coefficient for the y term?

Look at your proposed ODE again:

M*y" + N*y' + y = 0

What is the coefficient of the y term?

 

[ijd5000]

Look at your proposed ODE again:

M*y" + N*y' + y = 0

What is the coefficient of the y term?

i see it's 1 there, but isn't it possible for it to also be associated with some function?

also the same method doesn't seem to work for section (b), with only a y₁ ​solution?

*resolved