Differential equation ++

Hi.. I've got a few questions I'm stuck on and I'd appreciate some help..

In the previous questions I found out that :

u''(t)a+v''(t)b= (-9u(t)+\frac{27}{2})a+(-27v(t)-\frac{27}{2})b

which gave the equations:

u''(t) +9u(t)=\frac{27}{2}, u(0)=\frac{3}{2} u'(0)=\frac{1}{2}

v''(t) +27v(t)=\frac{27}{2}, v(0)=-\frac{1}{2} v'(0)=\frac{3}{2}

Ok.. So here is the question:

The general solution to the differential equation: z''(t)+kz(t)=a
is given by:

z(t)=\frac{a}{k}+Csin(\sqrt{k}t)+Dcos(\sqrt{k}t)

use this formula to find u(t) and v(t).

What I've tried so far:

Noticed that u''(t) and v''(t) is on the same form as z''(t) so:

u(t)=\frac{3}{2}+Csin(\sqrt{9}t)+Dcos(\sqrt{9}t)

v(t)=-\frac{1}{2}+Csin(\sqrt{27}t)+Dcos(\sqrt{27}t)

u(0)=\frac{3}{2}+Csin0+Dcos0

If what I've done so far is right, how do I find C and D? I end up with D and C = 0 which can't be right.. (also sorry if I have used wrong terminology and stuff somewhere, have to translate the question into english)

Reply 1

mathz

dont you have from u(0)=3/2 that D=0 then

u'(t)=3C cos(3t)

then u'(0)=1/2 gives

C=1/6

(edited 11 years ago)

Reply 2

Karaiyar

The other question I need help on:

x''(t)=Ax(t)+c

x(0) =

{1 \choose 2}

x'(0)=

{2 \choose -1}

x(t)=

{x(t) \choose y(t)}

A =

{-18, 9 \choose 9,-18}

c =

{0 \choose 27}

x(t) = u(t)a + v(t)b

find x(t) and y(t)..

Need to solve the other question before I can attempt this one, but how would I go about solving it?

Reply 3

Karaiyar

Original post by mathz

dont you have from u(0)=3/2 that D=0 then

u'(t)=3C cos(3t)

then u'(0)=1/2 gives

C=1/6

Thanks! I see what I did wrong now :smile:

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Differential equation ++

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