Inhomogenous D.e,,help needed!

• June 9th 2008, 07:28 PM
i_zz_y_ill
Inhomogenous D.e,,help needed!
par (iv) (B)* im stuck on, pls bare with me
q1 A solution is sought to te differential eqn

y(doubledot)+4(dot)+4y=2e^-2t
(i)find the complementary function
(ii)Explain why an expression of either of the fomrs ae^-2t or ate^-2t cannot be a particular integral of the D.e. State a correct form for the partcular integral and hence show the the general solution is:

y=(A+Bt+t^2)e^-2t
(iii)It is given that y=0 when t=1 and that y=0 when t=3. Calculate the constants A and B and write the particular solution in factorised form.
(iv)By reference to the previous answer, write dwn the particular solution in the cases
(A) y=0 when t=2 and y=0 when t=7
(B)*y=0 only when t=1

the answer in (iii) was (t-1)(t-3)e^-2t
I can't really see how this relates to (B).
I just put in y=0 and t=1 getting A+B+1=0 in which case A and B can be anything adding up to -1. The mark scheme i have however gives a mark for the answer being y=e^-2t(t-1)^2 in which case I manipulated this aswer and found that A would equal 1 and B would equal -2.(which contradicts what i thought) Obviously I am missing something either from 3 or something fundamental or this Question is very badly worded or ultimately wrong. Any inputs would be highly appreciated thanks!
• June 9th 2008, 08:01 PM
mr fantastic
Quote:

Originally Posted by i_zz_y_ill
par (iv) (B)* im stuck on, pls bare with me
q1 A solution is sought to te differential eqn

y(doubledot)+4(dot)+4y=2e^-2t
(i)find the complementary function
(ii)Explain why an expression of either of the fomrs ae^-2t or ate^-2t cannot be a particular integral of the D.e. State a correct form for the partcular integral and hence show the the general solution is:

y=(A+Bt+t^2)e^-2t
(iii)It is given that y=0 when t=1 and that y=0 when t=3. Calculate the constants A and B and write the particular solution in factorised form.
(iv)By reference to the previous answer, write dwn the particular solution in the cases
(A) y=0 when t=2 and y=0 when t=7
(B)*y=0 only when t=1

the answer in (iii) was (t-1)(t-3)e^-2t
I can't really see how this relates to (B).
I just put in y=0 and t=1 getting A+B+1=0 in which case A and B can be anything adding up to -1. The mark scheme i have however gives a mark for the answer being y=e^-2t(t-1)^2 in which case I manipulated this aswer and found that A would equal 1 and B would equal -2.(which contradicts what i thought) Obviously I am missing something either from 3 or something fundamental or this Question is very badly worded or ultimately wrong. Any inputs would be highly appreciated thanks!

First of all, I think part (iv) requires finding brand new values for A and B. So your starting point is

$y = (A + Bt + t^2) e^{-2t}$.

You require $(t - 1)^2 = A + Bt + t^2$ since this is the only way that y = 0 only when t = 1. So A = 1 and B = -2.

Incidentally, for the first bit you require $(t - 1)(t - 7) = A + Bt + t^2$ since this is the only way that y = 0 when t = 1 or t = 7 ......
• June 10th 2008, 04:51 AM
i_zz_y_ill
I can't just say i required (t-1)^2 i thought it could be like anything. When for (A) for instance two simultaneous eqns were calculated and A dn B were strictly limited, with (B) however heres only one imultaneos eqn A + B + 1 = 0

in which case there could be numerous answers to (B) do you see what i mean?
• June 10th 2008, 05:22 PM
mr fantastic
Quote:

Originally Posted by i_zz_y_ill
I can't just say i required (t-1)^2 i thought it could be like anything. When for (A) for instance two simultaneous eqns were calculated and A dn B were strictly limited, with (B) however heres only one imultaneos eqn A + B + 1 = 0

in which case there could be numerous answers to (B) do you see what i mean?

A particular solution has no arbitrary constants in it.

The only way $y = (A + Bt + t^2) e^{-2t}$ can be zero only for t = 1 is in the way that I've already said.